Class 12 Engineering Maths - Inverse Trigonometric Functions

$\sin^{-1} x + \cos^{-1} x =$ ________

$\sin^{-1} x + \cos^{-1} x = \cfrac{\pi}{2}$ when

$x \in [-1,1]$

If $xy < 1$ , $\tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } =$ _______.

Let

$\theta = \tan^{-1}x + \tan^{-1}y$

Taking

$\tan$ on both sides,

$\tan \theta$

$=\tan(\tan^{-1}x+\tan^{-1}y)\\=\cfrac{\tan(\tan^{-1}x)+\tan(\tan^{-1}y)}{1-\tan(\tan^{-1}x)\tan(\tan^{-1}y)}\\=\cfrac{x+y}{1-xy}$

Therefore,

$\theta = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$

Prove that $\tan ^{ -1 }{ (1) } +\tan ^{ -1 }{ (2) } +\tan ^{ -1 }{ (3) } =\pi$

Let

$\tan^{-1}(1)=x$

$\Rightarrow 1=\tan x$

Let

$\tan^{-1}(2)=y$

$\Rightarrow 2=\tan y$

Let

$\tan^{-1}(3)=z$

$\Rightarrow 3=\tan z$

$\tan \left ( x+y+z \right )=\cfrac{\tan x+\tan y+\tan z-\tan x.\tan y.\tan z}{1-\tan x.\tan y-\tan x.\tan z-\tan y.\tan z}$

$=\cfrac{1+2+3-1\times 2\times 3}{1-1\times 2-1\times 3-2\times 3}$

$=0$

$\Rightarrow x+y+z=\pi$

$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$

(

$x+y+z$ cannot be equal to zero, because

$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)$ will have some value greater than zero )

If $f : \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \rightarrow (-\infty, \infty)$ is defined by $f(x) = \tan x$ , then $f^{-1}( \sqrt{3}) =$

Given,

$f : \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \rightarrow (-\infty, \infty)$ is defined by

$f(x)=\tan x$ .

This function

$f(x)$ is one-one and onto in the given domain and range.

So it is invertible and and the inverse is defined by

$f^{-1}(y)=\tan^{-1}y$ for

$y\in (-\infty,\infty)$ .

$f^{-1}(\sqrt{3})=\tan^{-1}(\sqrt{3})=\dfrac{\pi}{3}$ . [ Since

$\tan 60^o=\sqrt{3}$ ].

If ${\sin ^{ - 1}}\sqrt {1 + x + {x^2}} + {\tan ^{ - 1}}\sqrt {x + {x^2}} = \dfrac{\pi }{2}$
$then\,x =$

${\tan ^{ - 1}}\sqrt {{x^2} + x} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \pi /2 \to (i)$

${\tan ^{ - 1}}\sqrt {{x^2} + x} = \theta$

$\sqrt {{x^2} + x} = \tan \theta$

$\sec \theta = \sqrt {1 + {{\tan }^2}\theta }$

$= \sqrt {1 + {x^2} + x}$

$\cos \theta = \dfrac{1}{{\sqrt {1 + {x^2} + x} }}$

${\cos ^{-1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2} + x} }}} \right) + {\sin ^{ - 1}}\left( {\sqrt {{x^2} + x + 1} } \right) = \pi /2$

${\cos ^{ - 1}}\alpha + {\sin ^{ - 1}}\alpha = \dfrac{\pi}{2}$ is possible when,

$\dfrac{1}{{\sqrt {1 + {x^2} + x} }} = \sqrt {{x^2} + x + 1}$

${x^2} + x + 1 = 1$

${x^2} + x = 0$

$x(x + 1) = 0$

$x = 0$ ,

$x = 1$

Simplify $\sin (\sin^{-1}(\dfrac {1}{5})+\cos^{-1}(x))=1$

$\begin{array}{l} \sin \left( { { { \sin }^{ -1 } }\dfrac { 1 }{ 5 } } \right) +{ \cos ^{ -1 } }x=1 \\ \Rightarrow \dfrac { 1 }{ 5 } +{ \cos ^{ -1 } }x=1 \\ \Rightarrow { \cos ^{ -1 } }x=1-\dfrac { 1 }{ 5 } \\ \Rightarrow { \cos ^{ -1 } }x=\dfrac { 4 }{ 5 } \\ x=\cos \dfrac { 4 }{ 5 } \end{array}$

Solve $\sin^{-1}\sqrt {x}$

$\int {{{\sin }^{ - 1}}\sqrt x }$

putting

$x=t^2$

$dx=2tdt$

$=2 \int {{{\sin }^{ - 1}}t \times tdt}$

Integrating by parts taking

$\sin ^{-1}t$ as first function.

$= 2\left[ {{{\sin }^{ - 1}}t \times \dfrac{{{t^2}}}{2} - \int {\dfrac{1}{{\sqrt {1 - {t^2}} }} \times \dfrac{{{t^2}}}{2}dt} } \right]$

$= {t^2}{\sin ^{ - 1}}t + \int {\frac{{1 - {t^2} + 1}}{{\sqrt {1 - {t^2}} }}dt} = {t^2}{\sin ^{ - 1}}t + \int {\sqrt {1 - {t^2}} dt} - \int {\dfrac{{dt}}{{\sqrt {1 - {t^2}} }}}$

$= {t^2}{\sin ^{ - 1}}t + \dfrac{t}{2}\sqrt {1 - {t^2}} + \dfrac{1}{2}{\sin ^{ - 1}}t - {\sin ^{ - 1}}t + C$

$= {t^2}{\sin ^{ - 1}}t + \dfrac{t}{2}\sqrt {1 - {t^2}} - \dfrac{1}{2}{\sin ^{ - 1}}t + C$

$= x{\sin ^{ - 1}}\sqrt x + \dfrac{{\sqrt x }}{2}\sqrt {1 - x} - \dfrac{1}{2}{\sin ^{ - 1}}\sqrt x + C$

If $\tan^{-1} \left (\dfrac{1}{2i^2}\right) = t$ , then $\tan t$ equals

$\sum_{i=1}^{\infty }\tan^{-1}\left ( \dfrac{1}{2i^2} \right )=t$

$\tan^{-1}\left ( \dfrac{1}{2i^2} \right )=t$

$\Rightarrow \tan t=\left ( \dfrac{1}{2i^2} \right )=\dfrac{-1}{2}$

Prove $\tan^{-1} \dfrac{3}{4} + \tan^{-1} \dfrac{3}{5} - \tan^{-1} \dfrac{8}{19} = \dfrac{\pi}{4}$ .

Formula to be used:

$\tan^{-1}a+\tan^{-1}b+\tan^{-1}c=\tan^{-1}\left [ \dfrac {a+b+c-abc}{1-ab-bc-ca} \right ]$

$\tan^{-1}\dfrac{3}{4}+\tan^{-1}\dfrac{3}{5}+\tan^{-1}\dfrac{8}{19}=\tan^{-1}\left [ \dfrac {\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{8}{19}-\dfrac{3}{4}\dfrac{3}{5}\dfrac{8}{19}}{1-\dfrac{3}{4}\dfrac{3}{5}-\dfrac{3}{5}\dfrac{8}{19}-\dfrac{8}{19}\dfrac{3}{4}} \right ]$

$\tan^{-1}\dfrac{3}{4}+\tan^{-1}\dfrac{3}{5}+\tan^{-1}\dfrac{8}{19}=\tan^{-1}\left [ \dfrac {\dfrac{425}{380}} {\dfrac{425}{380}} \right ]$

$\tan^{-1}\dfrac{3}{4}+\tan^{-1}\dfrac{3}{5}+\tan^{-1}\dfrac{8}{19}=\tan^{-1}1$

$\tan^{-1}\dfrac{3}{4}+\tan^{-1}\dfrac{3}{5}+\tan^{-1}\dfrac{8}{19}=\dfrac{\pi}{4}$

Evaluate : $\cos \left[ {{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\dfrac{\pi }{6} \right]$

Consider the given expression.

$\Rightarrow \cos \left[ {{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\dfrac{\pi }{6} \right]$

$\Rightarrow \cos \left[ {{\cos }^{-1}}\left( \cos \dfrac{5\pi }{6} \right)+\dfrac{\pi }{6} \right]$

$\Rightarrow \cos \left[ \dfrac{5\pi }{6}+\dfrac{\pi }{6} \right]$

$\Rightarrow \cos \left( \pi \right)$

$\Rightarrow -1$

Hence, this is the required result.

If ${\sin ^{ - 1}}x = \dfrac{\pi }{5}$ , for some $x \in \left( { - 1,1} \right)$ , then find the value of ${\cos ^{ - 1}}x$ .

We have,

${{\sin }^{-1}}x=\dfrac{\pi }{5}$ …….. (1)

We know that

${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$

From equation (1), we get

$\dfrac{\pi }{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2}$

${{\cos }^{-1}}x=\dfrac{\pi }{2}-\dfrac{\pi }{5}$

${{\cos }^{-1}}x=\dfrac{5\pi -2\pi }{10}$

${{\cos }^{-1}}x=\dfrac{3\pi }{10}$

Hence, this is the answer.

Solve ${\sin ^{ - 1}}\left( {\cos x} \right)$

Let,

$y={{\sin }^{-1}}\left( \cos x \right)$

$\Rightarrow y={{\sin }^{-1}}\sin \left( \dfrac{\pi }{2}-x \right)$

$\Rightarrow y=\dfrac{\pi }{2}-x$

Hence, this is the answer.

Evaluate: $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)+cosec^{-1}\dfrac{2}{\sqrt{3}}$ .

$\tan^{-1} \sqrt{3} - \sec^{-1} (-2) + cosec^{-1} \cfrac{2}{\sqrt3}$

$\cfrac{\pi}{3} -\cfrac{2\pi}{3}+\cfrac{\pi}{3}$

$\implies 0$

Prove that $\displaystyle \sin \cot^{-1} \tan \cos^{-1} x = \sin cosec^{-1} \cot \tan^{-1} x = x$ where $\displaystyle x \: \epsilon \: \left ( 0, \: 1 \right ]$

Let

$x=cos\theta$ . where

$cos\theta\epsilon(0,1]$
Hence

$\theta\epsilon [0,\dfrac{\pi}{2})\cup(\dfrac{3\pi}{2},2\pi]$ .
Then

$sin(cot^{-1}(tan(cos^{-1}(x))))$

$=sin(cot^{-1}(tan(cos^{-1}(cos\theta))))$

$=sin(cot^{-1}(tan\theta))$

$=sin(\dfrac{\pi}{2}-tan^{-1}(tan\theta))$

$=sin(\dfrac{\pi}{2}-\theta)$

$=cos\theta=x$ ...(i)
Similarly

$sin(cosec^{-1}(cot(tan^{-1}x)))$
Let

$x=tan\theta$ . where

$tan\theta\epsilon (0,1]$
Hence

$\theta\epsilon (0,\dfrac{\pi}{4}]\cup(\pi,\dfrac{5\pi}{4}]$
Then

$sin(cosec^{-1}(cot(tan^{-1}x)))$

$=sin(cosec^{-1}(cot(tan^{-1}tan\theta)))$

$=sin(cosec^{-1}(cot\theta))$

$=\dfrac{1}{cosec(cosec^{-1}(cot\theta))}$

$=\dfrac{1}{cot\theta}$

$=tan\theta$

$=x$ .

Write the function in the simplest form:
$\displaystyle { \tan }^{ -1 }\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } ,\left| x \right| >1$

$\displaystyle { \tan }^{ -1 }\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } ,\left| x \right| >1$

Put

$\displaystyle x=cosec\theta \Rightarrow \theta ={ cosec }^{ -1 }x$

$\displaystyle \therefore \quad { \tan }^{ -1 }\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } ={ \tan }^{ -1 }\frac { 1 }{ \sqrt { { cosec }^{ 2 }\theta -1 } }$

$\displaystyle ={ \tan }^{ -1 }\left( \frac { 1 }{ \cot { \theta } } \right) ={\tan }^{ -1 }\left( \tan { \theta } \right)$

$\displaystyle =\theta ={ cosec }^{ -1 }x$

$=\dfrac { \pi }{ 2 } -{ \sec }^{ -1 }x$ ...

$\left[\because { cosec }^{ -1 }x+{ \sec }^{ -1 }x=\dfrac { \pi }{ 2 } \right]$

Prove: $\displaystyle 2{ \tan }^{ -1 }\frac { 1 }{ 2 } +{ \tan }^{ -1 }\frac { 1 }{ 7 } ={ \tan }^{ -1 }\frac { 31 }{ 17 }$

$\text{L.H.S.}\displaystyle =2{ \tan }^{ -1 }\frac { 1 }{ 2 } +{ \tan }^{ -1 }\frac { 1 }{ 7 }$

$\displaystyle={ \tan }^{ -1 }\left(\frac { 2.\frac { 1 }{ 2 } }{ 1-{ \left( \frac { 1 }{ 2} \right) }^{ 2 } } \right)+{ \tan }^{ -1 }\frac { 1 }{ 7 }$

$= \tan ^{ -1 } \left(\dfrac { 1 }{ \frac 34} \right) + \tan ^{ -1 } \dfrac { 1 }{ 7 }$

$\displaystyle ={ \tan }^{ -1 }\frac { 4 }{ 3 } +{ \tan }^{ -1 }\frac { 1 }{ 7 }$

$\displaystyle={ \tan }^{ -1 } \left(\dfrac { \dfrac { 4 }{ 3 } +\dfrac { 1 }{ 7 } }{ 1-\dfrac {4 }{ 3 } .\dfrac { 1 }{ 7 } } \right)$

$\displaystyle ={ \tan }^{ -1 }\dfrac { \left( \dfrac { 28+3 }{ 21 } \right) }{ \left( \dfrac { 21-4 }{ 21 } \right) }$

$\displaystyle ={ \tan }^{ -1 }\frac { 31 }{ 17 } =\text{R.H.S.}$

Prove: $\displaystyle { \tan }^{ -1 }\frac { 2 }{ 11 } +{ \tan }^{ -1 }\frac { 7 }{ 24 } ={ \tan }^{ -1 }\frac { 1 }{ 2 }$

$\text{L.H.S.}\displaystyle ={ \tan }^{ -1 }\frac { 2 }{ 11 } +{ \tan }^{ -1 }\frac { 7 }{ 24 }$

$\displaystyle ={ \tan }^{ -1 }\dfrac { \dfrac { 2 }{ 11 } +\dfrac { 7 }{ 24 } }{ 1-\dfrac { 2 }{ 11 } .\dfrac { 7 }{ 24 } }$ , .....

$\left[\because { \tan }^{ -1 }x+{ \tan }^{ -1 }y={ \tan }^{ -1 }\dfrac { x+y }{ 1-xy } \right]$

$\displaystyle ={ \tan }^{ -1 }\dfrac { \dfrac { 48+77 }{ 11\times 24 } }{ \dfrac { 11\times 24-14 }{ 11\times 24 } }$

$\displaystyle ={ \tan }^{ -1 }\frac { 48+77 }{ 264-14 } ={ \tan }^{ -1 }\frac { 125 }{ 250 } ={ \tan }^{ -1 }\frac { 1 }{ 2 }$

$=\text{R.H.S.}$

If $\displaystyle { \tan }^{ -1 }\frac { x-1 }{ x-2 } +{ \tan }^{ -1 }\frac { x+1 }{ x+2 } =\frac {\pi }{ 4 }$ , then find the value of $x$ .

$\displaystyle { \tan }^{ -1 }\frac { x-1 }{ x-2 } +{ \tan }^{ -1 }\frac { x+1 }{ x+2 } =\frac { \pi }{ 4 }$ ...Given

$\displaystyle \Rightarrow { \tan }^{ -1 }\left[ \dfrac { \dfrac { x-1 }{ x-2 } +\dfrac {x+1 }{ x+2 } }{ 1-\left( \dfrac { x-1 }{ x-2 } \right) \left( \dfrac { x+1 }{ x+2 } \right) } \right] =\dfrac { \pi }{ 4 }$

$\displaystyle \Rightarrow { \tan }^{ -1 }\left[ \dfrac { \left( x-1 \right) \left( x+2 \right) +\left( x+1 \right) \left( x-2 \right) }{ \left( x+2 \right) \left( x-2 \right) -\left( x-1 \right) \left( x+1 \right) } \right] =\frac { \pi }{ 4 }$

$\displaystyle \Rightarrow { \tan }^{ -1 }\left[ \frac { { x }^{ 2 }+x-2+{ x }^{ 2}-x-2 }{ { x }^{ 2 }-4-{ x }^{ 2 }+1 } \right] =\frac { \pi }{ 4 }$

$\displaystyle \Rightarrow { \tan }^{ -1 }\left[ \frac { 2{ x }^{ 2 }-4 }{ -3 } \right] =\frac { \pi }{ 4 }$

$\displaystyle \Rightarrow \tan { \left[ { \tan }^{ -1 }\frac { 4-{ 2x }^{ 2 } }{ 3 } \right] } =\tan { \frac { \pi }{ 4 } }$

$\displaystyle \Rightarrow \frac { 4-{ 2x }^{ 2 } }{ 3 } =1$

$\displaystyle \Rightarrow 4-2{ x }^{ 2 }=3$

$\displaystyle \Rightarrow 2{ x }^{ 2 }=4-3=1$

$\displaystyle \Rightarrow x=\pm \frac { 1 }{ \sqrt { 2 } }$

If $\displaystyle \sin { \left( { \sin }^{ -1 }\frac { 1 }{ 5 } +{ \cos }^{ -1 }x \right) } =1$ , then find the value of $x$ .

$\displaystyle \sin { \left( { \sin }^{ -1 }\frac { 1 }{ 5 } +{ \cos }^{ -1 }x \right) } =1$

$\Rightarrow \displaystyle { \left( { \sin }^{ -1 }\frac { 1 }{ 5 } +{ \cos }^{ -1 }x \right) } =\sin^{-1}(1)=\frac{\pi}{2}$

$\Rightarrow \displaystyle \sin^{ -1 }\frac { 1 }{ 5 }=\frac{\pi}{2}-\cos^{-1}x=\sin^{-1}x$

$\displaystyle \therefore x=\frac{1}{5}$

Find the value of $\displaystyle \tan { \frac { 1 }{ 2 } } \left[ { \sin }^{ -1 }\frac { 2x }{ 1+{ x }^{ 2 } } +{ \cos }^{ -1 }\frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } } \right] ,\left| x \right| <1,y>0$ and $xy<1$

Let

$\displaystyle x=\tan { \theta } \Rightarrow \theta ={ \tan }^{ -1 }x$

$\displaystyle\therefore { \sin }^{ -1 }\frac { 2x }{ 1+{ x }^{ 2 } } ={ \sin }^{-1 }\left( \frac { 2\tan { \theta } }{ 1+{ \tan }^{ 2 }\theta } \right) ={ \sin }^{ -1 }\left( \sin2\theta \right) =2\theta =2{ \tan }^{ -1 }x$

Let

$\displaystyle y=\tan { \phi } \Rightarrow \phi ={ \tan }^{ -1 }y$

$\displaystyle\therefore { \cos }^{ -1 }\frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } }={ \cos }^{ -1 }\left( \frac { 1-{ \tan }^{ 2 }\phi }{ 1{ +\tan }^{ 2 }\phi } \right) ={ \cos }^{ -1 }\left( \cos { 2\phi } \right) =2\phi =2{ \tan }^{ -1 }y$

$\displaystyle \therefore \tan { \frac { 1}{ 2 } } \left[ { \sin }^{ -1 }\frac { 2x }{ 1+{ x }^{ 2 } } +{ \cos }^{-1 }\frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } } \right]$

$\displaystyle = \tan { \frac { 1 }{ 2 } } \left[ 2{ \tan }^{ -1 }x+2{ \tan }^{ -1 }y \right]$

$\displaystyle =\tan { \left[ { \tan }^{ -1 }x+{ \tan }^{ -1 }y \right] }$

$\displaystyle =\tan { \left[ { \tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] } =\frac { x+y }{ 1-xy }$

Prove: $\displaystyle \frac { 9\pi }{ 8 } -\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 1 }{ 3 }=\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 }$

$\displaystyle L.H.S.=\frac { 9\pi }{ 8 } -\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 1 }{ 3 }$

$\displaystyle =\frac { 9 }{ 4 } \left( \frac { \pi }{ 2 } -{ \sin }^{ -1 }\frac { 1 }{ 3 } \right)$

$\displaystyle=\frac { 9 }{ 4 } \left( { \cos }^{ -1 }\frac { 1 }{ 3 } \right) .....(1)$

Now, let

$\displaystyle { \cos }^{ -1 }\frac{ 1 }{ 3 }=x$ .

Then,

$\displaystyle \cos { x } =\frac { 1 }{ 3 } \Rightarrow \sin { x } =\sqrt { 1-{ \left( \frac { 1 }{ 3 } \right) }^{ 2 } } =\frac { 2\sqrt { 2 } }{ 3 }$

$\displaystyle \therefore x={ \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } \Rightarrow { \cos }^{ -1 }\frac { 1 }{ 3 } ={ \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 }$

$\displaystyle \therefore \text{L.H.S.}=\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } =\text{R.H.S.}$

If $A=\dfrac{1}{1}\cot^{-1}\left(\dfrac{1}{1}\right)+\dfrac{1}{2}\cot^{-1}\left(\dfrac{1}{2}\right)+\dfrac{1}{3}\cot^{-1}\left(\dfrac{1}{3}\right)$ and $B=1\cot^{-1}1+2\cot^{-1}2+3\cot^{-1}3$ then $|B-A|$ is equal to $\dfrac{a\pi}{b}+\dfrac{c}{d}\cot^{-1}3$ where $a, b, c, d \in N$ and are in their lowest form then $a + b + c+d$ equal to

Given, $A=\dfrac{1}{1}\cot^{-1}\left(\dfrac{1}{1}\right)+\dfrac{1}{2}\cot^{-1}\left(\dfrac{1}{2}\right)+\dfrac{1}{3}\cot^{-1}\left(\dfrac{1}{3}\right)$ and $B=1\cot^{-1}1+2\cot^{-1}2+3\cot^{-1}3$

We know that $\cot^{-1}\dfrac{1}{x}=\tan^{-1}x\ \forall\ x>0$

$\Rightarrow A = \tan^{-1}1 + \dfrac{1}{2}\tan^{-1}2 + \dfrac{1}{3}\tan^{-1}3$

Also, $\tan^{-1}1=\cot^{-1}1=\dfrac{\pi}{4}$

$\therefore |B-A| = |\cot^{-1}1 + 2\cot^{-1}2 + 3\cot^{-1}3 -(\tan^{-1}1 + \dfrac{1}{2}\tan^{-1}2 + \dfrac{1}{3}\tan^{-1}3)|$

$\therefore |B-A| = |2\cot^{-1}2 + 3\cot^{-1}3 -\dfrac{1}{2}\tan^{-1}2 - \dfrac{1}{3}\tan^{-1}3|$

Consider $\tan^{-1}2=x$

$\therefore \tan x=2$

$\therefore \sin x = 2\sqrt{1-\sin ^2x}$

$\therefore 5\sin ^2x=4$

$\therefore \sin x=\dfrac{2}{\sqrt{5}}\implies x=\dfrac{7\pi}{20}$

Also, $\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$

$\therefore |B-A| = |2(\dfrac{\pi}{2}-\dfrac{7\pi}{20}) + 3\cot^{-1}3 -\dfrac{7\pi}{40} - \dfrac{1}{3}(\dfrac{\pi}{2}-\cot^{-1}3)|$

$\therefore |B-A| = |\dfrac{3\pi}{10} -\dfrac{7\pi}{40} - \dfrac{\pi}{6}+ 3\cot^{-1}3 +\dfrac{1}{3}\cot^{-1}3|$

$\therefore |B-A| = |-\dfrac{\pi}{24}+ \dfrac{10}{3}\cot^{-1}3|$

$\therefore |B-A| = -\dfrac{\pi}{24}+ \dfrac{10}{3}\cot^{-1}3$ $\left(\because \cot^{-1}3>\dfrac{\pi}{24}\right)$

$\therefore a+b+c+d=-1+24+10+3=36$

Solve the equation: $\tan^{-1}x+2\cot^{-1}x=\displaystyle\frac{2\pi}{3}$ .

$\tan ^{-1}x+2\cot ^{-1}x=\displaystyle\frac{2\pi}{3}$

$\tan ^{-1}x+\cot ^{-1}+\cot ^{-1}x=\displaystyle\frac{2\pi}{3}$

$\displaystyle\frac{\pi}{2}+\cot ^{-1}x=\displaystyle\frac{2\pi}{3}$

$\cot ^{-1}x=\displaystyle\frac{2\pi}{3}-\frac{\pi}{2}$

$\cot ^{-1}x=\displaystyle\frac{\pi}{6}$

$x=\cot \displaystyle\frac{\pi}{6}$

$x=\sqrt{3}$ .

Write the simplest form of $\tan^{-1}\left (\dfrac {\cos x - \sin x}{\cos x + \sin x}\right ), 0 < x < \dfrac {\pi}{2}$ .

Given:

$\tan^{-1} \left (\dfrac {\cos x - \sin x}{\cos x + \sin x}\right ) = \tan^{-1}\left (\dfrac {1 - \dfrac {\sin x}{\cos x}}{1 + \dfrac {\sin x}{\cos x}}\right )$

$= \tan^{-1} \left (\dfrac {1 - \tan x}{1 + \tan x}\right )$

$= \tan^{-1} \left (\dfrac {\tan \dfrac {\pi}{4} - \tan x}{1 + \tan \dfrac {\pi}{4} \cdot \tan x}\right )$

$= \tan^{-1} \left (\tan \left (\dfrac {\pi}{4} - x\right )\right )$

$\therefore \tan^{-1} \left (\dfrac {\cos x - \sin x}{\cos x + \sin x}\right ) = \dfrac {\pi}{4} - x$ .

Write $\tan ^{ -1 }{ \left[ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right] } ,x\neq 0$ in the simplest form.

Let

$x=\tan { \theta }$

$\tan ^{ -1 }{ \left[ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right] } =\tan ^{ -1 }{ \left[ \cfrac { \sec { \theta } -1 }{ \cfrac { \sin { \theta } }{ \cos { \theta } } } \right] }$

$=\tan ^{ -1 }{ \left( \cfrac { 1-\cos { \theta } }{ \sin { \theta } } \right) }$

$=\tan ^{ -1 }{ \left( \cfrac { 2\sin ^{ 2 }{ \theta /2 } }{ 2\sin { \theta /2 } \cos { \theta /2 } } \right) }$

$=\tan ^{ -1 }{ \left( \tan { \theta /2 } \right) }$

$=\theta /2=\cfrac { 1 }{ 2 } \tan ^{ -1 }{ x }$

If $\sin \left (\sin^{-1} \dfrac {1}{5} + \cos^{-1} x\right ) = 1$ then find the value of $x$ .

Given

$\sin \left (\sin^{-1} \left (\dfrac {1}{5}\right ) + \cos^{-1}x\right ) = 1$

$\Rightarrow \sin^{-1} \left (\dfrac {1}{5}\right ) + \cos^{-1} x = \sin^{-1}(1)$

$\sin^{-1}\left (\dfrac {1}{5}\right ) + \cos^{-1}x = \dfrac {\pi}{2}$

$\therefore \cos^{-1} x = \dfrac {\pi}{2} - \sin^{-1} \left (\dfrac {1}{5}\right )$

$x = \cos \left (\dfrac {\pi}{2} - \sin^{-1} \left (\dfrac {1}{5}\right )\right )$

$x = \sin \left (\sin^{-1} \left (\dfrac {1}{5}\right )\right )$
or

$x = \dfrac {1}{5}$ .

Inverse circular functions,Principal values of ${ sin }^{ -1 }x,{ cos }^{ -1 }x,{ tan }^{ -1 }x$ .
${ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy }$ , $xy<1$
$\pi +{ tan }^{ -1 }\frac { x+y }{ 1-xy }$ , $xy>1$ .
(a) ${ cos }^{ -1 }\left( cos\frac { 7\pi }{ 6 } \right)$
(b) ${ cos }^{ -1 }\left( cos\frac { 4\pi }{ 3 } \right)$

(a)

${ cos }^{ -1 }cos\left( \dfrac { 7\pi }{ 6 } \right) =\dfrac { 7\pi }{ 6 }$ is wrong, as

$7\pi /6$ is greater than

$\pi$ .

Here

$\dfrac { 7\pi }{ 6 } =2\pi -\dfrac { 5\pi }{ 6 }$ so that

$\dfrac { 5\pi }{ 6 }$ lies between

$0$ and

$\pi$

$\therefore$

${ cos }^{ -1 }cos\left( \dfrac { 7\pi }{ 6 } \right) ={ cos }^{ -1 }cos\left( 2\pi -\dfrac { 5\pi }{ 6 } \right)$

$={ cos }^{ -1 }cos\left( \dfrac { 5\pi }{ 6 } \right) =\dfrac { 5\pi }{ 6 }$ .

(b)

${ cos }^{ -1 }cos\dfrac { 4\pi }{ 3 } ={ cos }^{ -1 }cos\left( 2\pi -\dfrac { 2\pi }{ 3 } \right)$

$={ cos }^{ -1 }cos\dfrac { 2\pi }{ 3 } =\dfrac { 2\pi }{ 3 }$ .

Inverse circular functions,Principal values of ${ sin }^{ -1 }x,{ cos }^{ -1 }x,{ tan }^{ -1 }x$ .
${ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy }$ , $xy<1$
$\pi +{ tan }^{ -1 }\frac { x+y }{ 1-xy }$ , $xy>1$ .
(a) ${ tan }^{ -1 }\frac { 2mn }{ { m }^{ 2 }-{ n }^{ 2 } } +{ tan }^{ -1 }\frac { 2pq }{ { p }^{ 2 }-{ q }^{ 2 } } ={ tan }^{ -1 }\frac { 2MN }{ { M }^{ 2 }-{ N }^{ 2 } }$
where $M=mp-nq,N=np+mq$
(b) $\frac { 2 }{ 3 } { tan }^{ -1 }\frac { 3m{ n }^{ 2 }-{ m }^{ 3 } }{ { n }^{ 3 }-3{ m }^{ 2 }n } +\frac { 2 }{ 3 } { tan }^{ -1 }\frac { 3p{ q }^{ 2 }-{ p }^{ 3 } }{ { q }^{ 3 }-3{ p }^{ 2 }q } ={ tan }^{ -1 }\frac { 2MN }{ { M }^{ 2 }-{ N }^{ 2 } }$
Where $M=-mp+nq,N=np+mq$ and all the letters are $+ive$ quantities.

(a) Dividing

${ T }_{ 1 },{ T }_{ 2 }$ and

${T}_{3}$ by

${m}^{2},{p}^{2}$ and

${M}^{2}$ respectively,

$2{ tan }^{ -1 }x={ tan }^{ -1 }\dfrac { 2x }{ 1-{ x }^{ 2 } }$

L.H.S.

$=2{ tan }^{ -1 }\dfrac { n }{ m } +2{ tan }^{ -1 }\dfrac { q }{ p }$

$=2{ tan }^{ -1 }\dfrac { pn+mq }{ pm-nq } =2{ tan }^{ -1 }\dfrac { N }{ M } =$ R.H.S.

(b) Dividing above and below by

${n}^{3}$ in

${T}_{1}$ and

${q}_{3}$ in

${T}_{2}$ and

${M}^{2}$ in

${T}_{3}$ , we get

$\dfrac { 2 }{ 3 } ,3{ tan }^{ -1 }\dfrac { m }{ n } +\dfrac { 2 }{ 3 } ,3{ tan }^{ -1 }\dfrac { p }{ q } =2{ tan }^{ -1 }\dfrac { N }{ M }$

L.H.S.

$=2\left[ { tan }^{ -1 }\dfrac { (m/n)+(p/q) }{ 1-(m/n)(p/q) } \right]$

$=2{ tan }^{ -1 }\dfrac { np+mq }{ nq-mp } =2{ tan }^{ -1 }\dfrac { N }{ M } =$ R.H.S.

Solve the equation $\tan^{-1} \sqrt{x^2+x}+\sin^{-1} \sqrt{x^2+x+1}=\dfrac{\pi}{2}$

$\tan^{-1}\sqrt{x^{2}+x}+\sin^{-1}\sqrt{x^{2}+x+1}=\dfrac{\pi }{2}$

$\tan^{-1}\sqrt{x^{2}+x}=\dfrac{\pi }{2}-\sin^{-1}\sqrt{x^{2}+x+1}$

$\tan^{-1}\sqrt{x^{2}+x}=\cos^{-1}\sqrt{x^{2}+x+1}$

Apply

$\tan^{-1}y=\cos^{-1}\dfrac{1}{\sqrt{1+y^{2}}}$ on R.H.S.

$\cos^{-1}\dfrac{1}{\sqrt{1+(\sqrt{x^{2}+x})^{2}}}=\cos^{-1}\sqrt{x^{2}+x+1}$

$\cos^{-1}\dfrac{1}{\sqrt{x^{2}+x+1}}=\cos^{-1}\sqrt{x^{2}+x+1}$

$\Rightarrow \dfrac{1}{\sqrt{x^{2}+x+1}}=\sqrt{x^{2}+x+1}$

$x^{2}+x+1=1$

$x^{2}+x=0$

$x(x+1)=0$

$x=0,-1$

If $\cot^{-1} \left( \frac{\sqrt{1+ \sin x}+ \sqrt{ 1- \sin x}}{\sqrt{1+\sin x}-\sqrt{1- \sin x}}\right) =\dfrac{x}{m}, x \in \left(0, \frac{\pi}{4}\right)$ .Find $m$

$\\0<x<(\frac{\pi}{4})\\hence\>0<(\frac{x}{2})<(\frac{\pi}{8})\\\>and\>so\>sin\>(\frac{x}{2})<cos(\frac{x}{2})\>\\\therefore\>\sqrt{1+sinx}=\sqrt{sin^2(\frac{x}{2})+cos^2(\frac{x}{2})+2sin(\frac{x}{2})cos(\frac{x}{2})}\\=sin(\frac{x}{2})+cos(\frac{x}{2})\\and\>\sqrt\>{1-sinx}=cos(\frac{x}{2})-sin(\frac{x}{2})\\\>hence\>given\>expression\>in\>question\>\\\>=cot^-1[(\frac{\sqrt{sin(\frac{x}{2})+cos(\frac{x}{2})+cos(\frac{x}{2})-sin(\frac{x}{2})}}{\sqrt{sin(\frac{x}{2})+cos(\frac{x}{2})-cos(\frac{x}{2})+sin(\frac{x}{2})}})]\\=cot^{-1}(cot(\frac{x}{2}))\\=(\frac{x}{2})\>hence\>m\>=2$

If $(Tan^{-1}x)^2+(Cot^{-1}x)^2=\dfrac{5\pi^2}{8},then$ x=

$(\tan^{-1}x)^2+(\dfrac{\pi}{2}-\tan^{-1}x)^2=\dfrac{5\pi^2}{8}$

Put

$\tan^{-1}x=c$

$\therefore 2c^2+\dfrac{\pi^2}{4}-\pi c-\dfrac{5\pi^2}{8}=0$

$\therefore 16c^2-8\pi c-3\pi^2=0$

$\therefore 16c^2-12\pi c+4\pi c-3\pi^2=0$

$(4c-3\pi)(4c+\pi)=0$

$c=\dfrac{3\pi}{4},\dfrac{-\pi}{4}$

$\tan^{-1}x=\dfrac{3\pi}{4},\dfrac{-\pi}{4}$

$x=\tan\dfrac{3\pi}{4},\tan\dfrac{-\pi}{4}$

$x=-1,1$

find the value of the following:
$(i) \ \sin^{-1}(\frac{-1}{2})$
$(ii) \ \cos^{-1}(\frac{\sqrt{3}}{2})$
$(iii) \ \text{cosec}^{-1}(2)$
$(iv) \ \tan^{-1} ({-\sqrt{3}})$
$(v) \ \cos^{-1}(\frac{-1}{2})$
$(vi) \tan^{-1} (-1)$

$\sin^{-1}(\cfrac{-1}{2})$

$=-\sin^{-1}(\cfrac{1}{2})$

$=\cfrac{-\pi}{6}$

ii)

$\cos^{-1}(\cfrac{\sqrt 3}{2})$ =

$\cfrac{\pi}{6}$

iii)

$\csc^{-1}(2)=\sin^{-1}(\cfrac{1}{2})=\cfrac{\pi}{6}$

iv)

$\tan^{-1}(-\sqrt 3)=-\tan^{-1}(\sqrt 3)=-\cfrac{\pi}{3}$

$\cos^{-1}(\cfrac{-1}{2})=\cfrac{2\pi}{3}$

vi)

$\tan^{-1}=\cfrac{-\pi}{4}$

Differentiate $\cos^{-1}(4x^{2} -3x); x \epsilon (\frac{1}{2}, 1)$

$\dfrac{d}{dx}\cos^{-1}(4x^{2} -3x)\\=\dfrac{-1}{\sqrt{1-(4x^{2} -3x)^2}}.\dfrac{d}{dx}(4x^2-3x)=\dfrac{-1}{\sqrt{1-(4x^{2} -3x)^2}}.(8x-3)$

Solve the following for $x:\tan^{-1}\left(\dfrac{x-2}{x-3}\right)+\tan^{-1}\left(\dfrac{x+2}{x+3}\right)=\dfrac{\pi}{4},|x|<1$

1314728_1035338_ans_57b3307b9bf14728a4941aae91b57bce.jpeg

Simplify : $\cot^{-1} [\sqrt{1 + x^2} - x]$

$\cot ^{-1}\left [ \sqrt{1+x^{2}} -x\right ]$

Let

$x=\cot\theta$

$\cot ^{-1}\left [ \sqrt{1+\cot^{2}\theta } -\cot\theta \right ]$

$=\cot ^{-1}\left [ \text{cosec} \theta -\cot\theta \right ]$

$=\cot ^{-1}\left [ \dfrac{1}{\sin \theta } -\dfrac{\cos \theta }{\sin \theta } \right ]$

$=\cot ^{-1}\left ( \dfrac{1-\cos \theta }{\sin \theta } \right )$

$=\cot ^{-1}\left ( \dfrac{2\sin ^{2}\dfrac{\theta }{2} }{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} } \right )$

$=\cot ^{-1}\left ( \tan\dfrac{\theta }{2} \right )$

$=\cot ^{-1}\left ( \cot\left ( \dfrac{\pi }{2}-\dfrac{\theta }{2} \right ) \right )$

$=\dfrac{\pi }{2}-\dfrac{\theta }{2}$

$=\dfrac{\pi }{2}-\dfrac{\cot^{-1}x }{2}$

Does $\sin^{-1} \left(\dfrac{2}{\sqrt{5}} \right) = \tan^{-1} (2)$ & please give value.

$\sin^{-1}\left ( \dfrac{2}{\sqrt{5}} \right )= x$

$\sin x=\dfrac{2}{\sqrt{5}}$

$\cos x=\dfrac{\sqrt{5-2^{2}}}{\sqrt{5}}$ =

$\dfrac{1}{\sqrt{5}}$

$\cos x=\dfrac{1}{\sqrt{5}}$

$\tan x=\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{1}{\sqrt{5}}}$

$\tan x=2$

$\tan^{-1}z=x$

$\therefore \sin\left ( \dfrac{2}{\sqrt{5}} \right )=\tan^{-1}(2)$

Find the domain and range of the real function $\cfrac { x+2 }{ { x }^{ 2 }-8x-4 }$

Let

$f\left( x \right) =\cfrac { x+2 }{ { x }^{ 2 }-8x-4 } =y\rightarrow 1$

${ x }^{ 2 }-8x-4=0$

$x=\cfrac { 8\pm \sqrt { 64+16 } }{ 2 } =\cfrac { 8\pm 4\sqrt { 5 } }{ 2 }$

$=4\pm 2\sqrt { 5 }$

$x=4\pm 2\sqrt { 5 } ,{ x }^{ 2 }-8x-4=0\quad \therefore f\left( x \right)$ is not defined

Domain of

$f(x)$ is

$R-\left\{ 4-2\sqrt { 5 } ,4+2\sqrt { 5 } \right\}$

$1\Rightarrow y\left( { x }^{ 2 }-8x-4 \right) =x+2$

$y{ x }^{ 2 }-(1+8y)x-\left( 4y+2 \right) =0$

As x is always real, Discriminant of above equation must be

$\ge 0$

${ (1+8y) }^{ 2 }+4y(4y+2)\ge 0$

$64{ y }^{ 2 }+16y+1+16{ y }^{ 2 }+8y\ge 0$

$80{ y }^{ 2 }+24y+1\ge 0$

Factorize above equation,

$y=\cfrac { -24\pm \sqrt { 526-320 } }{ 160 }$

$=\cfrac { -1 }{ 4 }$ or

$\cfrac { -1 }{ 20 }$

$\therefore \left( 4y+1 \right) \left( 20y+1 \right) \ge 0$

$\therefore y\le \cfrac { -1 }{ 20 }$ or

$y\ge \cfrac { -1 }{ 4 }$

$\therefore$ Range of

$f\left( x \right) =\left( -\infty ,\cfrac { -1 }{ 20 } \right] \bigcup { \left[ \cfrac { -1 }{ 4 } ,\infty \right) }$

Solve $y=\tan ^{ -1 }{ \cfrac { 5ax }{ a^{ 2 }-6{ x }^{ 2 } } }$ , find $\cfrac { dy }{ dx }$

$y=\tan ^{ -1 }{ \left( \cfrac { 5ax }{ { a }^{ 2 }-6{ x }^{ 2 } } \right) }$

$\tan { y } =\cfrac { 5ax }{ { a }^{ 2 }-6{ x }^{ 2 } } \quad \ast \sec ^{ 2 }{ y } -\tan ^{ 2 }{ y } =1$

$\left( \sec ^{ 2 }{ y } \right) \cfrac { dy }{ dx } =\cfrac { 5a\left( { a }^{ 2 }-6{ x }^{ 2 } \right) -\left( -12x \right) \left( 5ax \right) }{ { \left( { a }^{ 2 }-6{ x }^{ 2 } \right) }^{ 2 } }$

$\left[ 1+{ \left( \cfrac { 5ax }{ { a }^{ 2 }-6{ x }^{ 2 } } \right) }^{ 2 } \right] \cfrac { dy }{ dx } =\cfrac { 5{ a }^{ 3 }-30a{ x }^{ 2 }+60a{ x }^{ 2 } }{ { \left( { a }^{ 2 }-6{ x }^{ 2 } \right) }^{ 2 } }$

$\left[ \cfrac { { a }^{ 4 }+36{ x }^{ 4 }+13{ a }^{ 2 }{ x }^{ 2 } }{ { \left( { a }^{ 2 }-6{ x }^{ 2 } \right) }^{ 2 } } \right] \cfrac { dy }{ dx } =\cfrac { 5{ a }^{ 3 }+30a{ x }^{ 2 } }{ { \left( { a }^{ 2 }-6{ x }^{ 2 } \right) }^{ 2 } }$

$\left[ \left( { a }^{ 4 } \right) +36{ x }^{ 4 }+13{ a }^{ 2 }{ x }^{ 2 } \right] \cfrac { dy }{ dx } =5a\left( { a }^{ 2 }+6{ x }^{ 2 } \right)$

$\left( { a }^{ 2 }+9{ x }^{ 2 } \right) \left( { a }^{ 2 }+4{ x }^{ 2 } \right) \cfrac { dy }{ dx } =5a\left( { a }^{ 2 }+6{ x }^{ 2 } \right)$

$\therefore \cfrac { dy }{ dx } =\cfrac { 5a\left( { a }^{ 2 }+6{ x }^{ 2 } \right) }{ \left( { a }^{ 2 }+9{ x }^{ 2 } \right) \left( { a }^{ 2 }+4{ x }^{ 2 } \right) }$

$\cfrac { dy }{ dx } =\cfrac { 3a\left( { a }^{ 2 }+4{ x }^{ 2 } \right) +2a\left( { a }^{ 2 }+9{ x }^{ 2 } \right) }{ \left( { a }^{ 2 }+9{ x }^{ 2 } \right) \left( { a }^{ 2 }+4{ x }^{ 2 } \right) }$

$\therefore \cfrac { dy }{ dx } =\cfrac { 3a }{ { a }^{ 2 }+9{ x }^{ 2 } } +\cfrac { 2a }{ { a }^{ 2 }+4{ x }^{ 2 } }$

Calculating the principal value, find the value of $\sin\left[2\sin^{-1}\left(\dfrac{4}{5}\right)\right]$ .

We are given,

$sin \left [ 2xin^{-1}\left ( \dfrac{4}{5} \right ) \right ]$

Let

$sin^{-1}\left ( \dfrac{4}{5} \right ) = \theta$

where

$\theta \epsilon \left [ -\dfrac{\pi }{2}, \dfrac{\pi}{2} \right ]$

So,

$sin \theta = \dfrac{4}{5}$

now,

$cos\theta = \sqrt{cos^{2}\theta}$

$= \sqrt{1-sin^{2}\theta}$

$cos \theta = \sqrt{1-\left ( \dfrac{4}{5} \right )^2}$

$cos \theta = \sqrt{1-\dfrac{16}{25}}$

$cos \theta = \sqrt{\dfrac{9}{25}} = \pm \dfrac{3}{5}$

but

$\theta \epsilon \left [ -\pi/2, \pi/2 \right ]$ so,

$cos \theta = \dfrac{3}{5}$

$\therefore sin \left [ 2sin^{-1}\left ( \dfrac{4}{5} \right ) \right ]$

$= sin 2\theta$

$= 2sin \theta \,cos \theta$

$= 2\left ( \dfrac{4}{5} \right )\left ( \dfrac{3}{5} \right )$

$= \dfrac{24}{25}$

So,

$sin \left [ 2 sin^{-1} \dfrac{4}{5} \right ] = \dfrac{24}{25}$

1180647_1069989_ans_edd5770c9aba4fd0a8398412b9837e0b.jpg

Find $\sin ^{ -1 }{ \left( \cfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) }$ =

We know that

$\sin 75^0=\dfrac{\sqrt 3 +1}{2\sqrt 2}$

$\Rightarrow 75^0=\sin ^{-1}(\dfrac{\sqrt 3 +1}{2\sqrt 2})$

Therefore,

$\Rightarrow \sin ^{-1}(\dfrac{\sqrt 3 +1}{2\sqrt 2})=75^0$

Prove that ${\sin ^{ - 1}}\dfrac{3}{5} - {\sin ^{ - 1}}\dfrac{8}{{17}} = {\cos ^{ - 1}}\left( {\dfrac{{84}}{{85}}} \right)$

L.H.S

${{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{8}{12}$

We know that

${{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left\{ x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right\}$

Therefore,

$={{\sin }^{-1}}\left\{ \dfrac{3}{5}\sqrt{1-{{\left( \dfrac{8}{17} \right)}^{2}}}-\dfrac{8}{17}\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \right\}$

$={{\sin }^{-1}}\left\{ \dfrac{3}{5}\sqrt{1-\dfrac{64}{289}}-\dfrac{8}{17}\sqrt{1-\dfrac{9}{25}} \right\}$

$={{\sin }^{-1}}\left\{ \dfrac{3}{5}\sqrt{\dfrac{225}{289}}-\dfrac{8}{17}\sqrt{\dfrac{16}{25}} \right\}$

$={{\sin }^{-1}}\left\{ \dfrac{3}{5}\times \dfrac{15}{17}-\dfrac{8}{17}\times \dfrac{4}{5} \right\}$

$={{\sin }^{-1}}\left\{ \dfrac{45}{85}-\dfrac{32}{85} \right\}$

$={{\sin }^{-1}}\left\{ \dfrac{13}{85} \right\}$

$={{\cos }^{-1}}\left( \dfrac{84}{85} \right)$

Hence, proved.

$2{\tan ^{ - 1}}\left( {\frac{{1 + x}}{{1 - x}}} \right) + {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) =$

$2\tan ^{ -1 }{ \left( \dfrac { 1+x }{ 1-x } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) }$

$\tan ^{ -1 }{ \left( \dfrac { 1+x }{ 1-x } \right) } +\tan ^{ -1 }{ \left( \dfrac { 1+x }{ 1-x } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) }$

$\tan ^{ -1 }{ \left( \dfrac { \dfrac { 2+2x }{ 1-x } }{ 1-{ \left( \dfrac { 1+x }{ 1-x } \right) }^{ 2 } } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) }$

$\tan ^{ -1 }{ \left( \dfrac { 2+2x\left( 1-x \right) }{ -4x } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) }$

$-\tan ^{ -1 }{ \left( \dfrac { 1+{ x }^{ 2 } }{ 2x } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) }$

We know that,

$\tan ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 2x } \right) }$

$=\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) }$

$\Rightarrow -\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) } =0$

Prove
$4\tan^{-1}\left(\dfrac{1}{5}\right)-\tan^{-1}\left(\dfrac{1}{70}\right)+\tan^{-1}\left(\dfrac{1}{99}\right)=\dfrac{\pi}{4}$ .

To be proved :

$4 \tan^{-1}\dfrac{1}{5}-\tan^{-1}\left(\dfrac{1}{70}\right)+ \tan^{-1}\left(\dfrac{1}{99}\right)=\dfrac{\pi }{4}$

We know that

$2\tan^{-1}x= \tan^{-1}\left(\dfrac{2x}{1-x^{2}}\right)$

So,

$2 \tan^{-1}\dfrac{1}{5}=\tan^{-1}\left(\dfrac{2\times \dfrac{1}{5}}{1-\left(\dfrac{1}{5}\right)^{2}}\right)= \tan^{-1}\left(\dfrac{5}{12}\right)$

$LHS : 4 \tan^{-1}\dfrac{1}{5}+ \tan^{-1}\dfrac{1}{99}-\tan^{-1}\dfrac{1}{70}$

$\displaystyle =\left(2 \tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{70}\right)+\left(2 \tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{70}\right)+\left(2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{99}\right)$

$\displaystyle =\left(\tan^{-1}\frac{5}{12}-\tan^{-1}\frac{1}{70}\right)+\left(\tan^{-1}\frac{5}{12}+\tan^{-1}\frac{1}{99}\right)$

$\displaystyle= \tan^{-1}\left(\dfrac{\dfrac{5}{12}-\dfrac{1}{70}}{1+\left(\dfrac{5}{12}\right)\left(\dfrac{1}{70}\right)}\right)+\tan^{-1}\left(\dfrac{\dfrac{5}{12}+\dfrac{1}{99}}{1-\left(\dfrac{5}{12}\right)\left(\dfrac{1}{99}\right)}\right)$

$\left(\because \tan^{-1}x+\tan^{-1}y= \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)\right)$

$= \tan^{-1}\left(\dfrac{2}{5}\right)+\tan^{-1}\left(\dfrac{3}{7}\right)$

$=\tan^{-1}\left(\dfrac{\dfrac{2}{5}+\frac{3}{7}}{1-\dfrac{6}{35}}\right)= \tan^{-1}(1)$

$=\dfrac{\pi }{4}$

Find the value of $\sin\left(\dfrac{1}{2}\cot^{-1}\left(-\dfrac{3}{4}\right)\right)$ .

$\begin{matrix} \sin \left( { \frac { 1 }{ 2 } { { \cot }^{ -1 } }\left( { \frac { { -3 } }{ 4 } } \right) } \right) \\ let\, \, \, \frac { 1 }{ 2 } { \cot ^{ -1 } }\left( { \frac { { -3 } }{ 4 } } \right) =\theta \\ { \cot ^{ -1 } }\left( { \frac { { -3 } }{ 4 } } \right) =2\theta \\ \cot 2\theta =\frac { { -3 } }{ 4 } \\ \therefore \tan 2\theta =\frac { { -4 } }{ 3 } \\ Then\, \, \, \sin \left( \theta \right)=\dfrac{-2}{\sqrt13} \end{matrix}$

if $\int \sin^{-1}\left ( \dfrac{2x+2}{\sqrt{4x^2+8x+13}} \right )dx$ $=(x + 1)tan^{-1}\left ( \dfrac{2x+2}{3}\right )$ + $\lambda$ $ln(4x^2+8x+13)$ + $C$
then find the value of - 4 $\lambda$

77292

Given

$\sin ^ {-1} \left( \dfrac {2x+2}{\sqrt{4x^2+8x+13}}\right)$

$\sin ^ {-1} \left( \dfrac {2x+2}{\sqrt{(2x+2)^2+(3)^2}}\right)$

$\tan ^ {-1} \left( \dfrac {2x+2}{3}\right)$

$\displaystyle \int \tan ^ {-1} \left( \dfrac{2(x+1)}3\right) dx$

Put

$t=1+x \implies dt=dx$

$\displaystyle \int \tan ^ {-1} \left( \dfrac {2t}3\right)$

Integrating by parts

$\implies \displaystyle \tan ^ {-1} \dfrac {2t}3 \int dt-\int\left( \dfrac d{dt}\tan ^ {-1}\dfrac {2t}3 \times \int dt\right)$

$\implies \displaystyle t\tan ^ {-1} \dfrac {2t}3 -\int\left( \dfrac 1{1+\dfrac{4t^2}9} \times \dfrac 23 t \right) dt$

$\implies \displaystyle t\tan ^ {-1} \dfrac {2t}3 - \dfrac 68\int\left( \dfrac {8tdt}{9+{4t^2}} \right) dt$

$\implies \displaystyle t\tan ^ {-1} \dfrac {2t}3 - \dfrac 34 \ln (9+4t^2)+c$

$\implies \displaystyle (1+x)\tan ^ {-1} \dfrac {2x+2}3 - \dfrac 34 \ln (4x^2+4x+13)+c$

$\implies \lambda= \dfrac{-3}4$

$\implies -4\lambda =3$

If $0< \cos^{-1} x< 1$ and $1+\sin (\cos^{-1}x)+\sin^{2}(\cos^{-1}x)+\sin^{3} (\cos^{-1}x)+.........\infty =2,$ then find $2\sqrt{3}x.$

We have

$1+sin (cos^{-1}x)+sin^{2}(cos^{-1}x)+.........\infty =2$

This is an infinite

$GP$

$\Rightarrow \dfrac{1}{1-sin\left ( cos^{-1}x \right )}=2\Rightarrow \dfrac{1}{2}=1-sin \left ( cos^{-1}x \right )$

$\Rightarrow sin (cos^{-1}x)=\dfrac{1}{2}$

$\Rightarrow cos^{-1}x=\dfrac{\pi }{6} \therefore x=\dfrac{\sqrt{3}}{2} Ans.$

$\Rightarrow 2\sqrt{3}x=2\sqrt{3}\times\dfrac{\sqrt{3}}{2}=3$

Find the number of values of $x$ of the form $6n$ , where $n$ is an integer , in the domain of the function $f(x)=x\ln|x-1|+\displaystyle \frac{\sqrt{64-x^{2}}}{\sin x}$

Considering the domain, we get

$|x-1|>0$

$\Rightarrow x \neq 1$
And

$64-x^{2}\geq0$

$x^{2}\leq64$

$x\leq|8|$

$-8\leq x\leq 8$
Hence
domain is

$x\epsilon [-8,1)\cup(1,8]$
Also,

$sin x \neq 0$
Hence,

$x \neq k\pi$
Hence the number of the form

$6n$ are

$-6,6$
Hence there will be two numbers of the form 6n.
Hence answer is 2.

If $\displaystyle 0< \cos ^{-1}x< 1$ and $\displaystyle 1+\sin(\cos^{-1}x)+\sin^{2}(\cos^{-1}x)+\sin^{3}(\cos^{-1}x)+...infinity= 2$ then the value of $\displaystyle 12 x^{2}$ is

The above equation is an infinite G.P
Hence sum of the G.P will be

$S$

$=\dfrac{a}{1-r}$

$=\dfrac{1}{1-\sin(\cos^{-1}(x))}$ ...(i)

Let

$\cos^{-1}(x)=y$

$\therefore \cos y=x$

$\sin y=\pm\sqrt{1-x^{2}}$

$y=\sin^{-1}(\pm\sqrt{1-x^{2}})$

$=\pm \sin^{-1}(\sqrt{1-x^{2}})$
Substituting in the equation, gives us

$=\dfrac{1}{1\pm\sqrt{1-x^{2}}}$

$=2$

$2\pm2 \sqrt{1-x^{2}}=1$

$2-1=\pm2 \sqrt{1-x^{2}}$

$\dfrac{1}{2}=\pm\sqrt{1-x^{2}}$

Squaring both sides, we get

$\dfrac{1}{4}=1-x^{2}$

$x^{2}=\dfrac{3}{4}$

Hence,

$12x^{2}$

$=12\times\dfrac{3}{4}$

$=9$ .

If the domain of the function $\displaystyle f(x)= \sqrt{3\cos^{-1}(4x)-\pi}$ is [a,b] then the value of $\displaystyle (4a+64b)$ is

$3cos^{-1}(4x)-\pi$
Now the domain of

$cos^{-1}(x)=[-1,1]$
Similarly

$\Rightarrow$

$cos^{-1}(4x)\rightarrow[0,\pi]$

$x\rightarrow [\dfrac{-1}{4},\dfrac{-1}{4}]$
Now

$\sqrt{3cos^{-1}(4x)-\pi}\geq0$

$\Rightarrow 3cos^{-1}(4x)\geq\pi$

$\Rightarrow cos^{-1}(4x)\geq\dfrac{\pi}{3}$

$\Rightarrow 4x\leq\dfrac{1}{2}$

$x\leq\dfrac{1}{8}$
Hence domain is

$x\epsilon[\dfrac{-1}{4},\dfrac{1}{8}]$
Thus,

$4a+64b$

$=-1+8$

$=7$

If $\displaystyle x= \sin^{-1}(a^{6}+1)\cos^{-1}(a^{4}+1)-\tan^{-1}(a^{2}+1),a\epsilon R$ , then the value of $\displaystyle \sec^{2}x$ is

Let

$a=0$
We get

$x=sin^{-1}(1)cos^{-1}(1)-tan^{-1}(1)$

$=\dfrac{\pi}{2}.0-\dfrac{\pi}{4}$

$=\dfrac{-\pi}{4}$
Hence

$sec(x)=\sqrt{2}$

$\Rightarrow sec^{2}(x)=2$

If range of the function $\displaystyle f(x)= \sin^{-1}x+2\tan^{-1}x+x^{2}+4x+1$ is $[p,q]$ , then the value of $(p+q)$ is

Now the domain of

$sin^{-1}(x)$ is

$[-1,1]$
Hence the domain of the above function is

$[-1,1]$ since the domain of

$tan(x)$ is

$(-\infty,\infty)$ and the rest is a linear function.
Hence
Domain is

$[-1,1]$
Now

$f(1)=\dfrac{\pi}{2}+2\dfrac{\pi}{4}+1+4+1$

$=\pi+6$ .

$f(-1)=\dfrac{-\pi}{2}-2\dfrac{\pi}{4}+1-4+1$

$=-\pi-2$
Hence

$p+q$

$=\pi+6+(-\pi-2)$

$=6-2$

$=4$

Prove that

$2\tan^{-1} \dfrac {1}{2} + \tan^{-1}\dfrac {1}{7} = \tan^{-1} \dfrac {31}{17}$

Consider L.H.S

$=2{\tan}^{-1}{\left(\dfrac{1}{2}\right)}+{\tan}^{-1}{\left(\dfrac{1}{7}\right)}$

We know that,

$2{\tan}^{-1}{A}={\tan}^{-1}{\left(\dfrac{2A}{1-{A}^{2}}\right)}$

$={\tan}^{-1}{\left(\dfrac{2\times \dfrac{1}{2}}{1-{\left(\dfrac{1}{2}\right))}^{2}}\right)}+{\tan}^{-1}{\left(\dfrac{1}{7}\right)}$

$={\tan}^{-1}{\left(\dfrac{1}{1-\dfrac{1}{4}}\right)}+{\tan}^{-1}{\left(\dfrac{1}{7}\right)}$

$={\tan}^{-1}{\left(\dfrac{1}{\dfrac{4-1}{4}}\right)}+{\tan}^{-1}{\left(\dfrac{1}{7}\right)}$

$={\tan}^{-1}{\left(\dfrac{1}{\dfrac{3}{4}}\right)}+{\tan}^{-1}{\left(\dfrac{1}{7}\right)}$

$={\tan}^{-1}{\left(\dfrac{4}{3}\right)}+{\tan}^{-1}{\left(\dfrac{1}{7}\right)}$

We know that

${\tan}^{-1}{A}+{\tan}^{-1}{B}={\tan}^{-1}{\left(\dfrac{A+B}{1-AB}\right)}$

$={\tan}^{-1}{\left(\dfrac{\dfrac{4}{3}+\dfrac{1}{7}}{1-\dfrac{4}{3}\times\dfrac{1}{7}}\right)}$

$={\tan}^{-1}{\left(\dfrac{\dfrac{28+3}{21}}{1-\dfrac{4}{21}}\right)}$

$={\tan}^{-1}{\left(\dfrac{\dfrac{31}{21}}{\dfrac{21-4}{21}}\right)}$

$={\tan}^{-1}{\left(\dfrac{31}{17}\right)}=$ R.H.S

Hence proved.

Prove that $\tan^{-1} \left [\dfrac {\sqrt {1 + x} + \sqrt {1 - x}}{\sqrt {1 + x} - \sqrt {1 - x}}\right ] = \dfrac {\pi}{4} + \dfrac {1}{2} \cos^{-1} x, 0 < x < 1$

For this we have to put

$x= \cos 2\theta$

From LHS we get,

$\tan^{-1}\left (\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right )$

Put

$x=\cos 2\theta$

$\tan^{-1}(\dfrac{\sqrt{1+\cos2 \theta}-\sqrt{1-\cos2 \theta}}{\sqrt{1+\cos2 \theta}+\sqrt{1-\cos2 \theta}}) .......1$

Since we know that

$\cos2 \theta= 2\cos^2 \theta-1$ or

$1-2\sin^2 \theta$

Put this value in equation 1 then we get

$=\tan^{-1}(\dfrac{\sqrt{ 2\cos^2 \theta}-\sqrt{2\sin^2 \theta}}{{ \sqrt{2\cos^2 \theta}}-\sqrt{2\sin^2 \theta}})$

$\tan^{-1}(\dfrac{{\sqrt{2}\cos \theta-\sqrt2\sin \theta}}{{\sqrt{2}\cos \theta+\sqrt2\sin \theta}})$

$\tan^{-1}(\dfrac{\sqrt{2}}{\sqrt{2}})(\dfrac{{\cos \theta-\sin \theta}}{{\cos \theta+\sin \theta}})$

$=(\dfrac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta})$

dividing by

$\cos \theta$

$\tan{-1}(\dfrac{\dfrac{\cos \theta}{\cos \theta}-\dfrac{\sin \theta}{\cos \theta}}{\dfrac{\cos \theta}{\cos \theta}-\dfrac{\sin \theta}{\cos \theta}})$

$\tan^{-1}(\dfrac{1- \tan \theta}{1+\tan \theta})$

$\tan^{-1}(\dfrac{1- \tan \theta}{1+1.\tan \theta})$

$\tan^{-1}(\dfrac{1- \tan \theta}{1+\tan\dfrac{\pi}{4}\tan \theta})$ [since

$\tan\dfrac{\pi}{4}=1$ ]

By using

$\tan(x+y)= \dfrac{\tan x-\tan y}{1+\tan x.\tan y}$

Thus, Put this value

$x=\dfrac{\pi}{4}$ and

$y= \theta$

$\tan^{-1}\tan(\dfrac{\pi}{4}- \theta)$

$=\dfrac{\pi}{4}- \theta ......2$

\since we know that

$x=\cos2 \theta$

$\cos^{-1}=2 \theta$

$\dfrac{1}{2}\cos^{-1}x= \theta$

Put this value in equation

$2$ then we get

$=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x$

Hence RHS proved.

The number of real solutions of the equation
$\sin^{-1}\left(\displaystyle\sum^{\infty}_{i=1}x^{i+1}-x\displaystyle\sum^{\infty}_{i=1}\left(\displaystyle\dfrac{x}{2}\right)^i\right)=\dfrac{\pi}{2}-\cos^{-1}\left(\displaystyle\sum^{\infty}_{i=1}\left(-\displaystyle\frac{x}{2}\right)^i-\displaystyle\sum^{\infty}_{i=1}(-x)^i\right)$ lying in the interval $\left(-\displaystyle\frac{1}{2}, \frac{1}{2}\right)$ is

$\displaystyle\sum^{\infty}_{i=1}x^{i+1}=\frac{x^2}{1-x}$

$\displaystyle\sum^{\infty}_{i=1}\left(\displaystyle\frac{x}{2}\right)^i=\frac{x}{2-x}$

$\displaystyle\sum^{\infty}_{i=1}\left(-\displaystyle\frac{x}{2}\right)^i=\frac{-x}{2+x}$

$\displaystyle\sum^{\infty}_{i=1}(-x)^{i}=\frac{-x}{1+x}$
To have real solutions

$\displaystyle\sum^{\infty}_{i=1}x^{i+1}-x\displaystyle\sum^{\infty}_{i=1}\left(\displaystyle\frac{x}{2}\right)^i=\displaystyle\sum^{\infty}_{i=1}\left(\displaystyle\frac{-x}{2}\right)^i-\displaystyle\sum^{\infty}_{i=1}(-x)^i$

$\displaystyle\frac{x^2}{1-x}-\frac{x^2}{2-x}=\frac{-x}{2+x}+\frac{x}{1+x}$

$x(x^3+2x^2+5x-2)=0$

$\therefore x=0$ and let

$f(x)=x^3+2x^2+5x-2$

$f\left(\displaystyle\frac{1}{2}\right)\cdot f\left(\displaystyle -\frac{1}{2}\right) < 0$
Hence two solutions exist.

Using principal values, evaluate the following
$\cos^{-1}(\cos\dfrac{2\pi}{3})$ + $\sin^{-1}(\sin\dfrac{2\pi}{3})$

$\cos^{-1}(\cos \cfrac{2\pi}{3})+\sin^{-1}(\sin \cfrac{2\pi}{3})$

Figure 1 represents graph of

$\sin^{-1}\sin x$

Figure 2 represents graph of

$\cos^{-1}\cos x$

Hence,

$\cos^{-1}\cos\cfrac{2\pi}{3}+\sin^{-1}\sin\cfrac{2\pi}{3}$

$=\cfrac{2\pi}{3}+\pi-\cfrac{2\pi}{3}$

$=\cfrac{2\pi}{3}+\cfrac{\pi}{3}=\pi$

1016966_1018950_ans_b4224ab6349a4a838d35e961b9ee45b5.png

$\dfrac{d}{dx}\sin ^{-1}\left(2x\sqrt{1-x^2}\right)=.......|x|>\dfrac{1}{\sqrt{2}}$

Let

$y= \sin^{-1}(2x\sqrt {1-x^2}) \quad |x|>\dfrac{1}{\sqrt 2}$

Let

$x=\sin \theta$

$\Rightarrow 1-x^2 = 1-\sin^\theta = \cos^2 \theta$

$\Rightarrow y = \sin^{-1}(2\sin \theta \sqrt {\cos^2 \theta })$

$= \sin^{-1} (2 \sin \theta |\cos \theta |)$

$=\sin^{-1}(\sin 2\theta)$

Now

$|x|> \dfrac{1}{\sqrt 2}$

$\Rightarrow \theta \in \left[\dfrac{-\pi}{2}, \dfrac{-\pi}{4}\right)\cup \left(\dfrac{\pi}{4},\pi/2\right]$

$\Rightarrow 2\theta \in \left[-\pi, -\dfrac{\pi}{2}\right) \cup \left( \pi/2, \pi \right]$

$\Rightarrow y = \sin^{-1}(\sin 2 \theta ) = \begin{cases} -\pi -2\theta \quad \quad \quad \quad \quad 2\theta \in [-\pi ,\dfrac { -\pi }{ 2 } ) \\ +\pi -2\theta \quad \quad \quad \quad \quad 2\theta \in (\pi/2 ,\pi ] \end{cases}$

$\Rightarrow y= \begin{cases} -\pi -2\sin^{-1}x \quad \quad \quad \quad \quad x \in [-1 ,-\dfrac { 1 }{ \sqrt {2} } ) \\ \pi -2\sin^{-1}x \quad \quad \quad \quad \quad \quad x \in ( \dfrac{1}{\sqrt 2} ,1 ] \end{cases}$

$\Rightarrow \dfrac{dy}{dx} = -2\ \dfrac{1}{\sqrt {1-x^2}}$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{-2}{\sqrt {1-x^2}}$

Find the value of $\sin (\cfrac{1}{4}\sin^{-1}\cfrac{\sqrt 63}{8})$

$\sin (\cfrac{1}{4}\sin^{-1}\cfrac{\sqrt 63}{8})$

Let

$\cfrac{1}{4} \sin^{-1}\cfrac{\sqrt 63}{8}=\theta$

$\Rightarrow \sin 4\theta=\cfrac{\sqrt 63}{8}=2\sin 2\theta$

Prove that $\tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{13}-\tan^{-1}\dfrac{2}{9}=0$

Simplifying the LHS of ${\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{{13}} - {\tan ^{ - 1}}\frac{2}{9} = 0$

${\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{{13}} - {\tan ^{ - 1}}\frac{2}{9} = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{7} + \frac{1}{{13}}}}{{1 - \frac{1}{7} \times \frac{1}{{13}}}}} \right) - {\tan ^{ - 1}}\frac{2}{9}$

$= {\tan ^{ - 1}}\left( {\frac{{20}}{{90}}} \right) - {\tan ^{ - 1}}\left( {\frac{2}{9}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{2}{9}} \right) - {\tan ^{ - 1}}\left( {\frac{2}{9}} \right)$

$= 0$

$= {\rm{RHS}}$

Hence proved.

${\sin ^{ - 1}}\sqrt {1 + x + {x^2}} + {\tan ^{ - 1}}\sqrt {x + {x^2}} = \frac{\pi }{2}$
$then\,x =$

2041482_1048663_ans_1badae781dc648fbae1c72961dcad22d.jpg

$\tan^{-1}\left(\dfrac{5-x}{6x^2-5x-3}\right)$ .

$\begin{matrix} { \tan ^{ -1 } }\left( { \dfrac { { 3-x } }{ { 6{ x^{ 2 } }-5x-3 } } } \right) \\ compare\, \, it\, \, with\, , \, { \tan ^{ -1 } }\left( { \dfrac { { a-b } }{ { 1+ab } } } \right) \\ \Rightarrow 1+ab=6{ x^{ 2 } }-5x-3 \\ \Rightarrow 6{ x^{ 2 } }-5x-3=1+ab\, \, \, \Rightarrow 6{ x^{ 2 } }-5x-4=ab \\ \Rightarrow \left( { 3x-4 } \right) \left( { 2x+1 } \right) =ab \\ Also\, \, 5-x=2x+1-\left( { 3x-4 } \right) \\ i.e,\, \, \, a=2x+1\, \, \, \, \, \, \, \, \, \, b=3x-4 \\ { \tan ^{ -1 } }\left( { \dfrac { { a-b } }{ { 1+ab } } } \right) \, \, \, =\, \, { \tan ^{ -1 } }a+{ \tan ^{ -1 } }b \\ \Rightarrow \boxed { { { \tan }^{ -1 } }\left( { 2x+1 } \right) +{ { \tan }^{ -1 } }\left( { 3x-4 } \right) } \\ \end{matrix}$

$\cot^{-1} \dfrac {1 - x}{1 + x}$ .

Let

$x = \tan \theta ,\,\theta = {\tan ^{ - 1}}x$

$\begin{array}{l} ={ \cot ^{ - } }\left[ { \dfrac { { 1-\tan \theta } }{ { 1+\tan \theta } } } \right] \\ ={ \cot ^{ - } }\left[ { \tan \left( { \dfrac { \pi }{ 4 } -\theta } \right) } \right] \\ ={ \cot ^{ - } }\left[ { \cot \left( { \dfrac { \pi }{ 2 } -\dfrac { \pi }{ 4 } +\theta } \right) } \right] \\ ={ \cot ^{ - } }\cdot \cot \left[ { \dfrac { \pi }{ 4 } +\theta } \right] \\ =\dfrac { \pi }{ 4 } +\theta \\ =\dfrac { \pi }{ 4 } ={ \tan ^{ - } }x \end{array}$

$\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)+\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)$ .

$\begin{matrix} { \tan ^{ -1 } }\left( { \frac { { 4x } }{ { 1+5{ x^{ 2 } } } } } \right) +{ \tan ^{ -1 } }\left( { \frac { { 2+3x } }{ { 3-2x } } } \right) \\ \Rightarrow { \tan ^{ -1 } }\left( { \frac { { 5xd-x } }{ { 1+5x\cdot x } } } \right) +\, { \tan ^{ -1 } }\left( { \frac { { \frac { 2 }{ 3 } +x } }{ { 1-\frac { 2 }{ 3 } \cdot x } } } \right) \\ \Rightarrow { \tan ^{ -1 } }5x-{ \tan ^{ -1 } }x+{ \tan ^{ -1 } }\frac { 2 }{ 3 } +{ \tan ^{ -1 } }x \\ \Rightarrow { \tan ^{ -1 } }\left( { \frac { { 5x+\frac { 2 }{ 3 } } }{ { 1-\frac { { 10x } }{ 3 } } } } \right) \\ \Rightarrow { \tan ^{ -1\, } }\left( { \frac { { 15x+2 } }{ { 3-10x } } } \right) \\ \\ \\ \end{matrix}$

Find the value of $\sin{\left[\dfrac{1}{2}\cot^{-1}{\left(\dfrac{-3}{4}\right)}\right]}$ .

$\sin \left[ \dfrac{1}{2}\cot ^{-1}\left( -\dfrac{3}{4}\right)\right]$

Let

$\theta=\cot^{-1}\left( -\dfrac{3}{4}\right)$

$\Rightarrow \cot \theta =\dfrac{-3}{4}$

so,

$\cos \theta =\dfrac{-3}{5}$

$\Rightarrow \theta =\cos^{-1}\left( \dfrac{-3}{5}\right)$

$\therefore \sin \left[ \dfrac{1}{2}\cot^{-1}\left( -\dfrac{3}{4}\right)\right]=\sin\left[ \cos^{-1}\left( -\dfrac{3}{5}\right)\right].....\left(1\right)$

we know that :

$\Rightarrow \cos \theta =1-2\sin^2\left( \dfrac{\theta}{2}\right)$

$\Rightarrow \sin \dfrac{\theta}{2}=\sqrt{\dfrac{1-\cos\theta}{2}}$

using this we can simplify

$\left( 1\right)$ as

$\Rightarrow \sin \left[ \dfrac{\cos^{-1}\left(\dfrac{-3}{5}\right)}{2}\right]=\sqrt{\dfrac{1-\cos\left( \cos^{-1}\left(\dfrac{-3}{5}\right)\right)}{2}}$

$=\sqrt{\dfrac{1-\left(\dfrac{-3}{5}\right)}{2}}$

$=\sqrt{\dfrac{1+\dfrac{3}{5}}{2}}$

$=\sqrt{\dfrac{8}{10}}$

$=\sqrt{\dfrac{4}{5}}$

$\therefore \sin\left[ \dfrac{1}{2}\cot^{-1}\left(\dfrac{-3}{4}\right)\right] =\sqrt{\dfrac{4}{5}}$

Hence, the answer is

$\sqrt{\dfrac{4}{5}}.$

Inverse Trigonometric Functions - Class 12 Engineering Maths - Extra Questions

sin−1x+cos−1x=\sin^{-1} x + \cos^{-1} x = ________

If xy<1xy < 1, tan−1x+tan−1y=\tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } = _______.

Prove that tan−1(1)+tan−1(2)+tan−1(3)=π\tan ^{ -1 }{ (1) } +\tan ^{ -1 }{ (2) } +\tan ^{ -1 }{ (3) } =\pi

If f:(−π2,π2)→(−∞,∞)f : \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \rightarrow (-\infty, \infty) is defined by f(x)=tanxf(x) = \tan x, then f−1(√3)=f^{-1}( \sqrt{3}) =

If sin−1√1+x+x2+tan−1√x+x2=π2{\sin ^{ - 1}}\sqrt {1 + x + {x^2}} + {\tan ^{ - 1}}\sqrt {x + {x^2}} = \dfrac{\pi }{2}thenx=then\,x =

Simplify sin(sin−1(15)+cos−1(x))=1\sin (\sin^{-1}(\dfrac {1}{5})+\cos^{-1}(x))=1

Solve sin−1√x\sin^{-1}\sqrt {x}

If tan−1(12i2)=t \tan^{-1} \left (\dfrac{1}{2i^2}\right) = t , then tant\tan t equals

Prove tan−134+tan−135−tan−1819=π4\tan^{-1} \dfrac{3}{4} + \tan^{-1} \dfrac{3}{5} - \tan^{-1} \dfrac{8}{19} = \dfrac{\pi}{4}.

Evaluate : cos[cos−1(−√32)+π6]\cos \left[ {{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\dfrac{\pi }{6} \right]

If sin−1x=π5{\sin ^{ - 1}}x = \dfrac{\pi }{5}, for some x∈(−1,1)x \in \left( { - 1,1} \right), then find the value of cos−1x{\cos ^{ - 1}}x.

Solve sin−1(cosx){\sin ^{ - 1}}\left( {\cos x} \right)

Evaluate: tan−1√3−sec−1(−2)+cosec−12√3\tan^{-1}\sqrt{3}-\sec^{-1}(-2)+cosec^{-1}\dfrac{2}{\sqrt{3}}.

Prove that sincot−1tancos−1x=sincosec−1cottan−1x=x\displaystyle \sin \cot^{-1} \tan \cos^{-1} x = \sin cosec^{-1} \cot \tan^{-1} x = x where xϵ(0,1]\displaystyle x \: \epsilon \: \left ( 0, \: 1 \right ]

Write the function in the simplest form:tan−11√x2−1,|x|>1\displaystyle { \tan }^{ -1 }\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } ,\left| x \right| >1

Prove: 2tan−112+tan−117=tan−13117\displaystyle 2{ \tan }^{ -1 }\frac { 1 }{ 2 } +{ \tan }^{ -1 }\frac { 1 }{ 7 } ={ \tan }^{ -1 }\frac { 31 }{ 17 }

Prove: tan−1211+tan−1724=tan−112\displaystyle { \tan }^{ -1 }\frac { 2 }{ 11 } +{ \tan }^{ -1 }\frac { 7 }{ 24 } ={ \tan }^{ -1 }\frac { 1 }{ 2 }

If tan−1x−1x−2+tan−1x+1x+2=π4\displaystyle { \tan }^{ -1 }\frac { x-1 }{ x-2 } +{ \tan }^{ -1 }\frac { x+1 }{ x+2 } =\frac {\pi }{ 4 } , then find the value of xx.

If sin(sin−115+cos−1x)=1\displaystyle \sin { \left( { \sin }^{ -1 }\frac { 1 }{ 5 } +{ \cos }^{ -1 }x \right) } =1, then find the value of xx.

Find the value of tan12[sin−12x1+x2+cos−11−y21+y2],|x|<1,y>0\displaystyle \tan { \frac { 1 }{ 2 } } \left[ { \sin }^{ -1 }\frac { 2x }{ 1+{ x }^{ 2 } } +{ \cos }^{ -1 }\frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } } \right] ,\left| x \right| <1,y>0 and xy<1xy<1

Prove: 9π8−94sin−113=94sin−12√23\displaystyle \frac { 9\pi }{ 8 } -\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 1 }{ 3 }=\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 }

Solve the equation: tan−1x+2cot−1x=2π3\tan^{-1}x+2\cot^{-1}x=\displaystyle\frac{2\pi}{3}.

Write the simplest form of tan−1(cosx−sinxcosx+sinx),0<x<π2\tan^{-1}\left (\dfrac {\cos x - \sin x}{\cos x + \sin x}\right ), 0 < x < \dfrac {\pi}{2}.

Write tan−1[√1+x2−1x],x≠0\tan ^{ -1 }{ \left[ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right] } ,x\neq 0 in the simplest form.

If sin(sin−115+cos−1x)=1\sin \left (\sin^{-1} \dfrac {1}{5} + \cos^{-1} x\right ) = 1 then find the value of xx.

Solve the equation tan−1√x2+x+sin−1√x2+x+1=π2 \tan^{-1} \sqrt{x^2+x}+\sin^{-1} \sqrt{x^2+x+1}=\dfrac{\pi}{2}

If cot−1(√1+sinx+√1−sinx√1+sinx−√1−sinx)=xm,x∈(0,π4) \cot^{-1} \left( \frac{\sqrt{1+ \sin x}+ \sqrt{ 1- \sin x}}{\sqrt{1+\sin x}-\sqrt{1- \sin x}}\right) =\dfrac{x}{m}, x \in \left(0, \frac{\pi}{4}\right).Find mm

If (Tan−1x)2+(Cot−1x)2=5π28,then(Tan^{-1}x)^2+(Cot^{-1}x)^2=\dfrac{5\pi^2}{8},then x=

Differentiate cos−1(4x2−3x);xϵ(12,1)\cos^{-1}(4x^{2} -3x); x \epsilon (\frac{1}{2}, 1)

Solve the following for x:tan−1(x−2x−3)+tan−1(x+2x+3)=π4,|x|<1x:\tan^{-1}\left(\dfrac{x-2}{x-3}\right)+\tan^{-1}\left(\dfrac{x+2}{x+3}\right)=\dfrac{\pi}{4},|x|<1

Simplify : cot−1[√1+x2−x]\cot^{-1} [\sqrt{1 + x^2} - x]

Does sin−1(2√5)=tan−1(2)\sin^{-1} \left(\dfrac{2}{\sqrt{5}} \right) = \tan^{-1} (2) & please give value.

Find the domain and range of the real function x+2x2−8x−4\cfrac { x+2 }{ { x }^{ 2 }-8x-4 }

Solve y=tan−15axa2−6x2y=\tan ^{ -1 }{ \cfrac { 5ax }{ a^{ 2 }-6{ x }^{ 2 } } } , find dydx\cfrac { dy }{ dx }

Calculating the principal value, find the value of sin[2sin−1(45)]\sin\left[2\sin^{-1}\left(\dfrac{4}{5}\right)\right].

Find sin−1(√3+12√2)\sin ^{ -1 }{ \left( \cfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) } =

Prove that sin−135−sin−1817=cos−1(8485){\sin ^{ - 1}}\dfrac{3}{5} - {\sin ^{ - 1}}\dfrac{8}{{17}} = {\cos ^{ - 1}}\left( {\dfrac{{84}}{{85}}} \right)

2tan−1(1+x1−x)+sin−1(1−x21+x2)=2{\tan ^{ - 1}}\left( {\frac{{1 + x}}{{1 - x}}} \right) + {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) =

Prove 4tan−1(15)−tan−1(170)+tan−1(199)=π44\tan^{-1}\left(\dfrac{1}{5}\right)-\tan^{-1}\left(\dfrac{1}{70}\right)+\tan^{-1}\left(\dfrac{1}{99}\right)=\dfrac{\pi}{4}.

Find the value of \sin\left(\dfrac{1}{2}\cot^{-1}\left(-\dfrac{3}{4}\right)\right)\sin\left(\dfrac{1}{2}\cot^{-1}\left(-\dfrac{3}{4}\right)\right).

If 0< \cos^{-1} x< 10< \cos^{-1} x< 1 and 1+\sin (\cos^{-1}x)+\sin^{2}(\cos^{-1}x)+\sin^{3} (\cos^{-1}x)+.........\infty =2,1+\sin (\cos^{-1}x)+\sin^{2}(\cos^{-1}x)+\sin^{3} (\cos^{-1}x)+.........\infty =2, then find 2\sqrt{3}x.2\sqrt{3}x.

Find the number of values of xx of the form 6n6n, where nn is an integer , in the domain of the function f(x)=x\ln|x-1|+\displaystyle \frac{\sqrt{64-x^{2}}}{\sin x}f(x)=x\ln|x-1|+\displaystyle \frac{\sqrt{64-x^{2}}}{\sin x}

If the domain of the function \displaystyle f(x)= \sqrt{3\cos^{-1}(4x)-\pi} \displaystyle f(x)= \sqrt{3\cos^{-1}(4x)-\pi} is [a,b] then the value of \displaystyle (4a+64b) \displaystyle (4a+64b) is

If \displaystyle x= \sin^{-1}(a^{6}+1)\cos^{-1}(a^{4}+1)-\tan^{-1}(a^{2}+1),a\epsilon R\displaystyle x= \sin^{-1}(a^{6}+1)\cos^{-1}(a^{4}+1)-\tan^{-1}(a^{2}+1),a\epsilon R , then the value of \displaystyle \sec^{2}x\displaystyle \sec^{2}x is

If range of the function \displaystyle f(x)= \sin^{-1}x+2\tan^{-1}x+x^{2}+4x+1 \displaystyle f(x)= \sin^{-1}x+2\tan^{-1}x+x^{2}+4x+1 is [p,q][p,q], then the value of (p+q)(p+q) is

Prove that 2\tan^{-1} \dfrac {1}{2} + \tan^{-1}\dfrac {1}{7} = \tan^{-1} \dfrac {31}{17}2\tan^{-1} \dfrac {1}{2} + \tan^{-1}\dfrac {1}{7} = \tan^{-1} \dfrac {31}{17}

Using principal values, evaluate the following\cos^{-1}(\cos\dfrac{2\pi}{3})\cos^{-1}(\cos\dfrac{2\pi}{3})+ \sin^{-1}(\sin\dfrac{2\pi}{3})\sin^{-1}(\sin\dfrac{2\pi}{3})

\dfrac{d}{dx}\sin ^{-1}\left(2x\sqrt{1-x^2}\right)=.......|x|>\dfrac{1}{\sqrt{2}}\dfrac{d}{dx}\sin ^{-1}\left(2x\sqrt{1-x^2}\right)=.......|x|>\dfrac{1}{\sqrt{2}}

Find the value of \sin (\cfrac{1}{4}\sin^{-1}\cfrac{\sqrt 63}{8})\sin (\cfrac{1}{4}\sin^{-1}\cfrac{\sqrt 63}{8})

Prove that \tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{13}-\tan^{-1}\dfrac{2}{9}=0\tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{13}-\tan^{-1}\dfrac{2}{9}=0

{\sin ^{ - 1}}\sqrt {1 + x + {x^2}} + {\tan ^{ - 1}}\sqrt {x + {x^2}} = \frac{\pi }{2}{\sin ^{ - 1}}\sqrt {1 + x + {x^2}} + {\tan ^{ - 1}}\sqrt {x + {x^2}} = \frac{\pi }{2}then\,x = then\,x =

\tan^{-1}\left(\dfrac{5-x}{6x^2-5x-3}\right)\tan^{-1}\left(\dfrac{5-x}{6x^2-5x-3}\right).

\cot^{-1} \dfrac {1 - x}{1 + x}\cot^{-1} \dfrac {1 - x}{1 + x}.

\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)+\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)+\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right).

Find the value of \sin{\left[\dfrac{1}{2}\cot^{-1}{\left(\dfrac{-3}{4}\right)}\right]}\sin{\left[\dfrac{1}{2}\cot^{-1}{\left(\dfrac{-3}{4}\right)}\right]}.

Class 12 Engineering Maths Extra Questions

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$\sin^{-1} x + \cos^{-1} x =$ ________

If $xy < 1$ , $\tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } =$ _______.

Prove that $\tan ^{ -1 }{ (1) } +\tan ^{ -1 }{ (2) } +\tan ^{ -1 }{ (3) } =\pi$

If $f : \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \rightarrow (-\infty, \infty)$ is defined by $f(x) = \tan x$ , then $f^{-1}( \sqrt{3}) =$

If ${\sin ^{ - 1}}\sqrt {1 + x + {x^2}} + {\tan ^{ - 1}}\sqrt {x + {x^2}} = \dfrac{\pi }{2}$
$then\,x =$

Simplify $\sin (\sin^{-1}(\dfrac {1}{5})+\cos^{-1}(x))=1$

Solve $\sin^{-1}\sqrt {x}$

If $\tan^{-1} \left (\dfrac{1}{2i^2}\right) = t$ , then $\tan t$ equals

Prove $\tan^{-1} \dfrac{3}{4} + \tan^{-1} \dfrac{3}{5} - \tan^{-1} \dfrac{8}{19} = \dfrac{\pi}{4}$ .

Evaluate : $\cos \left[ {{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\dfrac{\pi }{6} \right]$

If ${\sin ^{ - 1}}x = \dfrac{\pi }{5}$ , for some $x \in \left( { - 1,1} \right)$ , then find the value of ${\cos ^{ - 1}}x$ .

Solve ${\sin ^{ - 1}}\left( {\cos x} \right)$

Evaluate: $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)+cosec^{-1}\dfrac{2}{\sqrt{3}}$ .

Prove that $\displaystyle \sin \cot^{-1} \tan \cos^{-1} x = \sin cosec^{-1} \cot \tan^{-1} x = x$ where $\displaystyle x \: \epsilon \: \left ( 0, \: 1 \right ]$

Write the function in the simplest form:
$\displaystyle { \tan }^{ -1 }\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } ,\left| x \right| >1$

Prove: $\displaystyle 2{ \tan }^{ -1 }\frac { 1 }{ 2 } +{ \tan }^{ -1 }\frac { 1 }{ 7 } ={ \tan }^{ -1 }\frac { 31 }{ 17 }$

Prove: $\displaystyle { \tan }^{ -1 }\frac { 2 }{ 11 } +{ \tan }^{ -1 }\frac { 7 }{ 24 } ={ \tan }^{ -1 }\frac { 1 }{ 2 }$

If $\displaystyle { \tan }^{ -1 }\frac { x-1 }{ x-2 } +{ \tan }^{ -1 }\frac { x+1 }{ x+2 } =\frac {\pi }{ 4 }$ , then find the value of $x$ .

If $\displaystyle \sin { \left( { \sin }^{ -1 }\frac { 1 }{ 5 } +{ \cos }^{ -1 }x \right) } =1$ , then find the value of $x$ .

Find the value of $\displaystyle \tan { \frac { 1 }{ 2 } } \left[ { \sin }^{ -1 }\frac { 2x }{ 1+{ x }^{ 2 } } +{ \cos }^{ -1 }\frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } } \right] ,\left| x \right| <1,y>0$ and $xy<1$

Prove: $\displaystyle \frac { 9\pi }{ 8 } -\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 1 }{ 3 }=\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 }$

Solve the equation: $\tan^{-1}x+2\cot^{-1}x=\displaystyle\frac{2\pi}{3}$ .

Write the simplest form of $\tan^{-1}\left (\dfrac {\cos x - \sin x}{\cos x + \sin x}\right ), 0 < x < \dfrac {\pi}{2}$ .

Write $\tan ^{ -1 }{ \left[ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right] } ,x\neq 0$ in the simplest form.

If $\sin \left (\sin^{-1} \dfrac {1}{5} + \cos^{-1} x\right ) = 1$ then find the value of $x$ .

Solve the equation $\tan^{-1} \sqrt{x^2+x}+\sin^{-1} \sqrt{x^2+x+1}=\dfrac{\pi}{2}$

If $\cot^{-1} \left( \frac{\sqrt{1+ \sin x}+ \sqrt{ 1- \sin x}}{\sqrt{1+\sin x}-\sqrt{1- \sin x}}\right) =\dfrac{x}{m}, x \in \left(0, \frac{\pi}{4}\right)$ .Find $m$

If $(Tan^{-1}x)^2+(Cot^{-1}x)^2=\dfrac{5\pi^2}{8},then$ x=

Differentiate $\cos^{-1}(4x^{2} -3x); x \epsilon (\frac{1}{2}, 1)$

Solve the following for $x:\tan^{-1}\left(\dfrac{x-2}{x-3}\right)+\tan^{-1}\left(\dfrac{x+2}{x+3}\right)=\dfrac{\pi}{4},|x|<1$

Simplify : $\cot^{-1} [\sqrt{1 + x^2} - x]$

Does $\sin^{-1} \left(\dfrac{2}{\sqrt{5}} \right) = \tan^{-1} (2)$ & please give value.

Find the domain and range of the real function $\cfrac { x+2 }{ { x }^{ 2 }-8x-4 }$

Solve $y=\tan ^{ -1 }{ \cfrac { 5ax }{ a^{ 2 }-6{ x }^{ 2 } } }$ , find $\cfrac { dy }{ dx }$

Calculating the principal value, find the value of $\sin\left[2\sin^{-1}\left(\dfrac{4}{5}\right)\right]$ .

Find $\sin ^{ -1 }{ \left( \cfrac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) }$ =

Prove that ${\sin ^{ - 1}}\dfrac{3}{5} - {\sin ^{ - 1}}\dfrac{8}{{17}} = {\cos ^{ - 1}}\left( {\dfrac{{84}}{{85}}} \right)$

$2{\tan ^{ - 1}}\left( {\frac{{1 + x}}{{1 - x}}} \right) + {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) =$

Prove
$4\tan^{-1}\left(\dfrac{1}{5}\right)-\tan^{-1}\left(\dfrac{1}{70}\right)+\tan^{-1}\left(\dfrac{1}{99}\right)=\dfrac{\pi}{4}$ .

Find the value of $\sin\left(\dfrac{1}{2}\cot^{-1}\left(-\dfrac{3}{4}\right)\right)$ .

If $0< \cos^{-1} x< 1$ and $1+\sin (\cos^{-1}x)+\sin^{2}(\cos^{-1}x)+\sin^{3} (\cos^{-1}x)+.........\infty =2,$ then find $2\sqrt{3}x.$

Find the number of values of $x$ of the form $6n$ , where $n$ is an integer , in the domain of the function $f(x)=x\ln|x-1|+\displaystyle \frac{\sqrt{64-x^{2}}}{\sin x}$

If the domain of the function $\displaystyle f(x)= \sqrt{3\cos^{-1}(4x)-\pi}$ is [a,b] then the value of $\displaystyle (4a+64b)$ is

If $\displaystyle x= \sin^{-1}(a^{6}+1)\cos^{-1}(a^{4}+1)-\tan^{-1}(a^{2}+1),a\epsilon R$ , then the value of $\displaystyle \sec^{2}x$ is

If range of the function $\displaystyle f(x)= \sin^{-1}x+2\tan^{-1}x+x^{2}+4x+1$ is $[p,q]$ , then the value of $(p+q)$ is

Prove that

$2\tan^{-1} \dfrac {1}{2} + \tan^{-1}\dfrac {1}{7} = \tan^{-1} \dfrac {31}{17}$

Using principal values, evaluate the following
$\cos^{-1}(\cos\dfrac{2\pi}{3})$ + $\sin^{-1}(\sin\dfrac{2\pi}{3})$

$\dfrac{d}{dx}\sin ^{-1}\left(2x\sqrt{1-x^2}\right)=.......|x|>\dfrac{1}{\sqrt{2}}$

Find the value of $\sin (\cfrac{1}{4}\sin^{-1}\cfrac{\sqrt 63}{8})$

Prove that $\tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{13}-\tan^{-1}\dfrac{2}{9}=0$

${\sin ^{ - 1}}\sqrt {1 + x + {x^2}} + {\tan ^{ - 1}}\sqrt {x + {x^2}} = \frac{\pi }{2}$
$then\,x =$

$\tan^{-1}\left(\dfrac{5-x}{6x^2-5x-3}\right)$ .

$\cot^{-1} \dfrac {1 - x}{1 + x}$ .

$\tan^{-1}\left(\dfrac{4x}{1+5x^2}\right)+\tan^{-1}\left(\dfrac{2+3x}{3-2x}\right)$ .

Find the value of $\sin{\left[\dfrac{1}{2}\cot^{-1}{\left(\dfrac{-3}{4}\right)}\right]}$ .