If $$sin^{-1}(\dfrac{1}{3}) + sin^{-1}(\dfrac{2}{3}) = sin^{-1} x$$, then x is equal to

$$\dfrac{5\sqrt{2}-4\sqrt{5}}{9}$$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 5

$f ( x ) = \sqrt { \sec ^ { - 1 } \left( \frac { 2 - | x | } { 4 } \right) } ,$ then the domain of

$f ( x )$ is ______________.

0%
$[ - 2,2 ]$
0%
$[ - 6,6 ]$
0%
$( - \infty , - 6 ] \cup [ 6 , \infty )$
0%
None Of These

$3{\tan ^{ - 1}}\left( {\frac{1}{{2 + \sqrt 3 }}} \right) - {\tan ^{ - 1}}\frac{1}{x} = {\tan ^{ - 1}}\frac{1}{3}$ then

$x =$

0%
$1$
0%
$2$
0%
$3$
0%
$\sqrt 3$

Explanation

Let

$y=\text{tan}^{-1} \bigg(\dfrac{1}{2+\sqrt{3}}\bigg)\implies \tan y=\dfrac{1}{2+\sqrt{3}}$

$\tan 3 y=\dfrac{3\tan y-\tan^3 y}{1+3\tan^2 y}=\dfrac{\frac{1}{2+\sqrt{3}}(3-\frac{1}{(2+\sqrt{3})^2})}{1+\frac{3}{(2+\sqrt{3})^2}}=\dfrac{\frac{1}{2+\sqrt{3}}(20+12\sqrt{3})}{4+4\sqrt{3}}=\dfrac{20+12\sqrt{3}}{20+12\sqrt{3}}=1$

$\implies 3\text{tan}^{-1}\bigg(\dfrac{1}{2+\sqrt{3}}\bigg)=\text{tan}^{-1}(1)$

$\text{tan}^{-1}\bigg(\dfrac{1}{x}\bigg)=\text{tan}^{-1}(1)-\text{tan}^{-1}\dfrac{1}{3}=\text{tan}^{-1}\dfrac{1-\dfrac{1}{3}}{1+\dfrac{1}{3}}=\dfrac{1}{2}$

$\implies x=2$

$\displaystyle \sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \left(\dfrac { { 2 }^{ r-1 } }{ 1+{ 2 }^{ 2r-1 } } \right)$ is equal to :

0%
${ tan }^{ -1 }({ 2 }^{ n })$
0%
${ tan }^{ -1 }({ 2 }^{ n })-\frac { \pi }{ 4 }$
0%
${ tan }^{ -1 }({ 2 }^{ n+1 })$
0%
${ tan }^{ -1 }({ 2 }^{ n+1 })-\frac { \pi }{ 4 }$

Explanation

$\displaystyle\sum ^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^{r-1}}{1+2^{2 r-1}}\bigg)=\sum^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^{r-1}(2-1)}{1+2^r.2^{r-1}}\bigg)$

$=\displaystyle\sum^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^r-2^{r-1}}{1+2^r.2^{r-1}}\bigg)$

$=\displaystyle\sum^{n}_{r=1}\text{tan}^{-1}{2^r}-\text{tan}^{-1}(2^{r-1})$

$=\text{tan}^{-1} 2-\text{tan}^{-1}1+\text{tan}^{-1}2^2-\text{tan}^{-1}2^1+\cdots+\text{tan}^{-1}{2^n}-\text{tan}^{-1}2^{n-1}$

$=\text{tan}^{-1}(2^n)-\dfrac{\pi}{4}$

For

$0<x<1$ the value of

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

0%
$0$
0%
$\pi$
0%
$-\pi$
0%
none of these

Explanation

We've for

$|x|\le 1$ ,

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

It also holds in this case.

So for

$0<x<1$ we've,

$\cos^{-1}x+\cos^{-1} (-x)$

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

$=\pi$ .

The value of

$\sin\ h(\cos\ h^{-1}x)$ is

0%
$\sqrt{x^{2}+1}$
0%
$1/\sqrt{x^{2}+1}$
0%
$\sqrt{x^{2}-1}$
0%
$none of these$

The value of

$\displaystyle \sum_{r=0}^{\infty}{ tan}^{1}\left(\dfrac{1}{1+r+{r}^{2}}\right)$ is equal to

0%
$\dfrac { \pi } { 4 }$
0%
$- \dfrac { \pi } { 2 }$
0%
$\pi$
0%
$\dfrac { \pi } { 2 }$

Explanation

given,

$\displaystyle\sum^{\infty}_{r=0}\tan^{-1}\left(\dfrac{1}{1+r+r^2}\right)$

Consider

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

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

G.E

$=\displaystyle\sum^{\infty}_{r=0}\tan^{-1}(r+1)-\tan^{-1}(r)$

$=\tan^{-1}1-\tan^{-1}0+\tan^{-1}2-\tan^{-1}1+...…+\tan^{-1}\infty$

$=\tan^{-1}\infty -\tan^{-1}0$

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

$=\dfrac{\pi}{2}$ .

The value of

${ sin }^{ -1 }(sin12)+{ cos }^{ -1 }(cos12)$ is equal to :

0%
$Zero$
0%
$24-2\pi$
0%
$4\pi -24$
0%
None of these

Explanation

Given,

$\sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)$

The principle values,

$\sin ^{-1}x\in \left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ],\forall x\in [-1,1]$

$\cos ^{-1}x\in \left [0,\pi \right ],\forall x\in [-1,1]$

$\therefore \sin ^{-1}(\sin 12)\neq 12\notin \left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]$

$\therefore \cos ^{-1}(\cos 12)\neq 12\notin \left [ 0,\pi \right ]$

$\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)$

$=\sin ^{-1}(\sin (12-4\pi ))+\cos ^{-1}(\cos (4\pi -12))$

$=(12-4\pi ) + (4\pi -12)$

$=0$

$\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)=0$

The simplified form of

${ cos }^{ -1 }\left( \frac { 3 }{ 5 } { cos }x+\frac { 4 }{ 5 } { sin }x \right)$ is :

0%
${ tan }^{ -1 }\frac { 4 }{ 3 } -x$
0%
${ tan }^{ -1 }\frac { 1 }{ 3 } -x$
0%
${ tan }^{ -1 }\frac { 4 }{ 3 } +x$
0%
None of these

$\displaystyle \sum^{\infty}_{r=1}\tan^{-1}\left(\dfrac {3}{r^{2}-r+9}\right)$ is -

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {\pi}{12}$

$If\,\cos \left( {2{{\sin }^{ - 1}}x} \right) = \frac{1}{9},\,then\,\,x\,\,is\,\,equal\,\,to$

0%
$Only\,\frac{2}{3}\,$
0%
$Only - \frac{2}{{3\,}}$
0%
$\,\frac{2}{3}, - \frac{2}{3}$
0%
$\frac{1}{3}$

${ cos }^{ -1 }\dfrac { 3 }{ 5 } -{ sin }^{ -1 }\dfrac { 4 }{ 5 } ={ cos }^{ -1 }x$ , then x is equal to -

0%
0
0%
1
0%
-1
0%
None of these

Explanation

Given,

$\cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5}=\cos ^{-1}x$

apply

$\cos$ on both sides, we get,

$\cos \left ( \cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5} \right )=\cos (\cos ^{-1}x)$

$\mathrm{Use\:the\:following\:identity}:\quad \cos \left(s+t\right)=\cos \left(s\right)\cos \left(t\right)-\sin \left(s\right)\sin \left(t\right)$

$\cos \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\cos \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)-\sin \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\sin \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)=x$

$\dfrac{3}{5}\left ( \sqrt{1-\left (\dfrac{4}{5} \right )^2} \right )-\dfrac{4}{5}\left ( \sqrt{1-\left (\dfrac{3}{5} \right )^2} \right )=x$

$\dfrac{3}{5}\times \dfrac{3}{5}-\dfrac{4}{5}\times \dfrac{4}{5}=x$

$\dfrac{9}{25}-\dfrac{16}{25}=x$

$\therefore x=-\dfrac{7}{25}$

Number of solution of the equation

$\tan^{-1}\left(\dfrac {1}{x-1}+\dfrac {1}{x-2}+\dfrac {1}{x-3}+\dfrac {1}{x-4} \right)+\cos^{-1}(x)=\dfrac {3\pi}{4}-\sin^{-1}(x)$ are

0%
$0$
0%
$1$
0%
$2$
0%
$3$

The range of the function

$f(x)=\ell n\ (\sin^{-1}(x^{2}+x))$ is

0%
$\left[-\ell n \left(\sin^{-1}\dfrac {1}{4}\right),\ell n \dfrac {\pi}{4}\right]$
0%
$\left[-\ell n \dfrac {\pi}{4},\ell n \dfrac {\pi}{2}\right]$
0%
$\left(0,\ell n \dfrac {\pi}{2}\right]$
0%
$\left(-\infty ,\ell n \dfrac {\pi}{2}\right]$

The trigonometric equation

$\sin^{-1}x=2\sin^{-1}a$ , has a solution for-

0%
$\left|a\right|\le\dfrac{1}{\sqrt{2}}$
0%
$\dfrac{1}{2}<\left|a\right|<\dfrac{1}{\sqrt{2}}$
0%
$all\ real\ values\ of\ a$
0%
$\left|a\right|<\dfrac{1}{2}$

Explanation

The trigonometric equation

$\text{sin}^{-1} x=2\text{sin}^{-1} a$ will have a solution if

$2\text{sin}^{-1} a\in \bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigg]$

$(\because\text{sin}^{-1} x\in [-\frac{\pi}{2},\frac{\pi}{2}])$

$\implies \text{sin}^{-1} a\in \bigg[-\dfrac{\pi}{4},\dfrac{\pi}{4}\bigg]$

$\implies a\in \bigg[-\sin \dfrac{\pi}{4},\sin \dfrac{\pi}{4}\bigg]$

$\implies a\in \bigg[-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\bigg]$

$\implies |a|\le \dfrac{1}{\sqrt{2}}$

$f(x)=\cos^{-1}\left(\dfrac {\sqrt {2x^{2}+1}}{x^{2}+1}\right)$ , then range of

$f(x)$ is

0%
$[0,\pi]$
0%
$\left(0,\dfrac {\pi}{4}\right]$
0%
$\left(0,\dfrac {\pi}{3}\right]$
0%
$\left[0,\dfrac {\pi}{2}\right)$

Explanation

Given,

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

Range of

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

$0< f(x)\leq1$

$\cos^{-1}(1)\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)<\cos^{-1}(0)$

$0\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)< \dfrac{\pi}{2}$

Therefore the range is,

$0\leq f(x)< \dfrac{\pi}{2}$

Number of solution(s) to the equation

${\cos ^{ - 1}}x + {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = \dfrac{\pi }{6}\,$ is/are

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

Hence,

option

$(B)$ is correct answer.

1380779_1283290_ans_8fcbababf0de4892a5f785ac2967cb57.jpg

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

0%
-1
0%
1
0%
0
0%
None of these

Explanation

As we know that

$\text{sec}^{-1}(-2)=\dfrac{2\pi}{3}$

$\therefore \cos \bigg(\dfrac{\pi}{3}+\text{sec}^{-1}(-2)\bigg)=\cos \bigg(\dfrac{\pi}{3}+\dfrac{2\pi}{3}\bigg)=\cos \pi=-1$

The value of

$\cot\left(cosec^{-1}\dfrac{5}{3}+\tan^{-1}\dfrac{2}{3}\right)$ is equal to-

0%
$\dfrac{6}{17}$
0%
$\dfrac{3}{17}$
0%
$\dfrac{4}{17}$
0%
$\dfrac{5}{17}$

$\int _{ 0 }^{ \pi }{ \left[ cotx \right] dx,where\left[ \cdot \right] }$ denotes the greatest integer function, is equal to:

0%
1
0%
-1
0%
$-\dfrac { \pi }{ 2 }$
0%
$\dfrac { \pi }{ 2 }$

${ sec\quad h }^{ -1 }\left( sin\quad \theta \right) =$

0%
$log\left( tan\frac { \theta }{ 2 } \right)$
0%
$log\left( sin\frac { \theta }{ 2 } \right)$
0%
$log\left( cos\frac { \theta }{ 2 } \right)$
0%
$log\left( cot\frac { \theta }{ 2 } \right)$

$\cos^{ -1 }\left[\cos\left( -\frac { 17 }{ 15 } \pi \right) \right]$ is equal to

0%
$-\frac { 17\pi }{ 15 }$
0%
$\frac { 17\pi }{ 15 }$
0%
$\frac { 13\pi }{ 15 }$
0%
$\frac { -2\pi }{ 15 }$

Find

$\displaystyle \int x.\sin xdx$

0%
$\sin x+x\cos x=C$
0%
$\sin x-x\cos x+C$
0%
$\sin x+x\sin x+C$
0%
$\sin x-x\sin x+C$

$A=\tan^{1-}\left(\dfrac {x\sqrt {3}}{2k-x}\right)$ and

$B=\tan^{-1}\left(\dfrac {2x-k}{k\sqrt {3}}\right)$ then

$A.B=$

0%
$0^{o}$
0%
$\pi /6$
0%
$\pi /4$
0%
$\pi /3$

if x

$>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right)$

0%
${ log }_{ e }\left( 2x \right)$
0%
${ log }_{ e }x$
0%
${ log }_{ e }\left( 3x \right)$
0%
${ log }_{ e }\left( 5x \right)$

$if\quad x>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right)$

0%
${ log }_{ e }\left( 2x \right)$
0%
${ log }_{ e }x$
0%
${ log }_{ e }\left( 3x \right)$
0%
${ log }_{ e }\left( 5x \right)$

$\cos^{-1}x-\cos^{-1}(\dfrac {y}{2})=\alpha$

$ax^{2}-4xy\cos \alpha +y^{2}=$

0%
$-4\sin^{2}\alpha$
0%
$4\sin^{2}\alpha$
0%
$4$
0%
$2\sin 2\alpha$

$\cot^{-1}{x}+\tan^{-1}\left (\dfrac{1}{3}\right)=\dfrac{\pi}{2}$ , then

$x$ will be

0%
$1$
0%
$3$
0%
$\dfrac {1}{3}$
0%
None of these

Explanation

Given

$\text{cot}^{-1} x+\text{tan}^{-1}\bigg(\dfrac{1}{3}\bigg)=\dfrac{\pi}{2}$

$\implies \text{cot}^{-1} x=\dfrac{\pi}{2}-\text{tan}^{-1} \bigg(\dfrac{1}{3}\bigg)$

$\implies \text{cot}^{-1} x=\text{cot}^{-1}\bigg(\dfrac{1}{3}\bigg)$

$\implies x=\dfrac{1}{3}$

$x={ \sin }^{ -1 }(\sin10)$ and

$y={ \cos }^{ -1 }(\cos10)$ , then find

$y - x$ .

0%
$\pi$
0%
$7\pi$
0%
$0$
0%
$10$

Explanation

Given,

$x=sin^{-1}(sin 10)$

$x=3\pi -10$

Also,

$y=cos^{-1}(cos 10)$

$y=4\pi - 10$

Therefore,

$y-x=4\pi-10-3\pi+10 = \pi$

State true or false.

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

0%
True
0%
False

Explanation

As we know that

For

$y\in [-1,1],\text{sin}^{-1} y+\text{cos}^{-1} y=\dfrac{\pi}{2}$

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

So the relation is

$\text{True}$

${ \sin }^{ -1 }\left (\dfrac { 3 }{ 5 }\right )+{ \cos }^{ -1 }\left (\dfrac { 12 }{ 13 }\right )={ \sin }^{ -1 }\left (\dfrac { 56 }{ 65 } \right)$

0%
True
0%
False

$\cos ^{ -1 }{ \left\{ \dfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\dfrac { { x }^{ 2 } }{ 4 } } \right\} =\cos ^{ -1 }{ \dfrac { x }{ 2 } -\cos ^{ -1 }{ x } } }$ holds for:

0%
$|x|\le 1$
0%
$x\in R$
0%
$0\le x\le 1$
0%
$-1\le x\le 0$

Explanation

$clearly, x/2\epsilon[-1,1]$ &

$x\epsilon[-1,1]$

$x\epsilon [-2,2]$ &

$x\epsilon[-1,1]$

Intersection of above sells is [-1,1]

Thus

$,[\left | x \right |\leq 1]$

1174662_1347184_ans_7bb4c25a92a4450ca14a7954651e835c.jpg

$\frac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}=$

0%
4
0%
3
0%
2
0%
1

Explanation

$\textbf{Step-1: Rewrite the Numerator observing Denominator}$

$\textbf{& apply multiple - angles of trigonometric function.}$

$\text{We have,}$

$\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\$

$\text{.....(i)}$

$\text{Now,}\\$

$\cos^{-1} (1- 2 (\dfrac{2}{7})^2)$

$\text{Put}$

$\sin \theta = \dfrac{2}{7}$

$\therefore$

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

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

$\boldsymbol{[\cos 2 \theta = 1 - 2 \sin^{2} \theta]}$

$= 2\theta$

$\textbf{Step-2: Use the above results & get the required unknown.}$

$\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\$

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

$[\textbf{using (i)}]$

$\textbf{Hence, option - C is the answer}$

The value of

$tan(\frac { 1 }{ 2 } { cos }^{ -1 }(\frac { \sqrt { 5 } }{ 3 } ))$ is

0%
$\frac { 3+\sqrt { 5 } }{ 2 }$
0%
$3-\sqrt { 5 }$
0%
$\frac { 1 }{ 2 } (3-\sqrt { 5 } )$
0%
none of these

Explanation

Given,

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

Let,

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

$\Rightarrow \cos x=\dfrac{\sqrt{5}}{3}$

$\therefore y=\tan \dfrac{1}{2}x$

$y=\tan \dfrac{x}{2}$

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

$=\sqrt{\dfrac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}}$

$=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}$

rationalizing the factor, we get,

$y=\dfrac{3-\sqrt{5}}{2}$

$\sin^{-1}\left(\dfrac{4}{5}\right)+\sin^{-1}\left(\dfrac{7}{25}\right)=\sin^{-1}\left(\dfrac{117}{125}\right)$

0%
True
0%
False

If tan (x + y) = 33 and x =

${ tan }^{ -1 }3$ , then y will be

0%
0.3
0%
${ tan }^{ -1 }(1.3)$
0%
${ tan }^{ -1 }(0.3)$
0%
${ tan }^{ -1 }(\frac { 1 }{ 18 } )$

Explanation

Given,

$\tan (x+y)=33$

$x=\tan ^{-1}3$

$\Rightarrow \tan x=3$

Formula,

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

$33=\dfrac{3+\tan y}{1-(3)\tan y}$

$33(1-3\tan y)=3+\tan y$

$33-99\tan y=3+\tan y$

$30=100\tan y$

$\tan y=\dfrac{30}{100}=0.3$

$\therefore y=\tan ^{-1}0.3$

$sin^{-1}(\dfrac{1}{3}) + sin^{-1}(\dfrac{2}{3}) = sin^{-1} x$ , then x is equal to

0%
$0$
0%
$\dfrac{\sqrt{5}+4\sqrt{2}}{9}$
0%
$\dfrac{5\sqrt{2}-4\sqrt{5}}{9}$
0%
$\dfrac{\pi}{2}$

Explanation

$\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{2}{3}\\\sin ^{-1}\left\{\dfrac{1}{3}\sqrt{1-\left(\dfrac{2}{3}\right)^2}+\dfrac{2}{3}\sqrt{1-\left(\dfrac{1}{3}\right)^2}\right\}\\\sin^{-1}\left\{\dfrac{1}{3}\left(\dfrac{\sqrt5}{3}\right)+\dfrac{2}{3}\left(\dfrac{\sqrt8}{3}\right)\right\}\\\sin^{-1}\left\{\dfrac{\sqrt5+4\sqrt2}{9}\right\}\\x=\dfrac{\sqrt5+4\sqrt2}{9}$

The value of

$\sin ^{-1}(\sin 5\frac {\pi}{3})=$

0%
$-\frac {\pi}3$
0%
$\frac {\pi}3$
0%
$\frac {4\pi}3$
0%
$\frac {3\pi}3$

Explanation

Given,

$\sin ^{-1}\left (\sin 5\dfrac{\pi }{3} \right )$

Let

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

$\sin \theta =\left (\sin 5\dfrac{\pi }{3} \right )=\sin \left ( 2\pi - \dfrac{\pi }{3}\right )$

$\Rightarrow \sin \theta =-\sin \dfrac{\pi }{3}$

$\therefore \theta =-\dfrac{\pi }{3}$

The value of

$\displaystyle sec\left [ sin^{-1}\left (sin \dfrac{50\pi }{9} \right ) + cos^{-1}cos\left ( \dfrac{31\pi }{9} \right ) \right ]$ is equal to

0%
$sec\dfrac{10\pi}{9}$
0%
$sec \ 9{\pi}$
0%
$-1$
0%
$1$

Explanation

Given,

$\sec \left [ \sin ^{-1}\left ( \sin \dfrac {50\pi }{9} \right ) +\cos ^{-1}\left ( \cos \dfrac {31\pi }{9} \right ) \right ]$

$=\sec \left [ \dfrac {50\pi }{9} + \dfrac {31\pi }{9}\right ]$

$=\sec \dfrac {81\pi }{9}$

$=\sec 9\pi$

$\tan { ^{ -1 }\left( 3/5 \right) } +\tan { ^{ -1 }\left( 1/4 \right) } =$

0%
$0$
0%
$\pi /4$
0%
$3n/4$
0%
None of these

Explanation

given,

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

$={\tan}^{-1}{\left(\dfrac{\frac{3}{5}+\frac{1}{4}}{1-\frac{3}{5}\times\frac{1}{4}}\right)}$

$={\tan}^{-1}{\left(\dfrac{\frac{12+5}{20}}{1-\frac{3}{20}}\right)}$

$={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{20-3}{20}}\right)}$

$={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{17}{20}}\right)}$

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

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

$\tan { ^{ -1 }\left( \tan { 3\pi /4 } \right) = }$

0%
$5\pi /4$
0%
$\pi /4$
0%
$-\pi /4$
0%
None of these

Explanation

Let

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

$\Rightarrow \tan{y}=\tan{\left(\pi-\dfrac{\pi}{4}\right)}$

$\Rightarrow \tan{y}=\tan{\dfrac{\pi}{4}}$

$\therefore y=\dfrac{\pi}{4},\pi+\dfrac{\pi}{4}$

$\Rightarrow y=\dfrac{\pi}{4},\dfrac{4\pi+\pi}{4}$

$\therefore y=\dfrac{\pi}{4},\dfrac{5\pi}{4}$

The value of

$\sin^{-1}(\cos (\log_{2}(4\alpha -44)))$ is

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{2}$
0%
$0$
0%
$1$

$\tan (2\cos ^{-1}\frac 35)=$ _____

0%
$\frac 83$
0%
$\frac {24}{25}$
0%
$\frac 7{25}$
0%
$\frac {-24}7$

Explanation

Given,

$\tan \left(2\cos ^{-1}\left(\frac{3}{5}\right)\right)$

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

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

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

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

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

$=\dfrac{2\cdot \frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}}{1-\frac{4^2}{3^2}}$

$=\dfrac{\frac{8}{3}}{1-\frac{4^2}{3^2}}$

$=\dfrac{8}{3\left(1-\frac{16}{9}\right)}$

$=-\dfrac{8}{\frac{7}{3}}$

$=-\dfrac{24}{7}$

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

0%
True
0%
False

The solution set of the equation

$2 cos^{ -1 } x = cot^{ -1 } \left(\dfrac { 2x^{ 2 } - 1 }{ 2x \sqrt { 1 x^{ 2 } } }\right)$ is

0%
$\left(0, 1\right)$
0%
$\left(-1, 1\right)-{ 0 }$
0%
$\left(-1, 0\right)$
0%
$\left[-1, 1\right]$

${ \cos }^{ -1 }\left[ \cos \left( 2{ \cot }^{ -1 }\left( \sqrt { 2 } -1 \right) \right) \right]$ is equal to

0%
$\sqrt{2}-1$
0%
$1-\sqrt{2}$
0%
$\pi/4$
0%
$3\pi/4$

The number of solutions of the equation

$3\cos^{-1}x-\pi x-\dfrac {\pi}{2}=0$

0%
$0$
0%
$1$
0%
$2$
0%
$infinite$

$\alpha =\cos^{-1}\left(\dfrac{3}{5}\right),\beta=\tan^{-1}\left(\dfrac{1}{3}\right)$ where

$0<\alpha,beta <\dfrac{\pi}{2},$ then

$\alpha -\beta$ is equal to :

0%
$\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$
0%
$\tan^{-1}\left(\dfrac{9}{14}\right)$
0%
$\cos^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$
0%
$\tan^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$

Explanation

$\cos \alpha=\dfrac{3}{5},\tan \beta =\dfrac{1}{5}$

$\Rightarrow \tan \alpha =\dfrac{4}{3}$

$\Rightarrow tan (\alpha-\beta)=\dfrac{\dfrac{4}{3}-\dfrac{1}{3}}{1+\dfrac{4}{3}.\dfrac{1}{3}}=\dfrac{9}{3}$

$\Rightarrow \sin(\alpha-\beta)=\dfrac{9}{5\sqrt{10}}$

$\Rightarrow \alpha-\beta =\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$

$\;\sin {\;^{ - 1}}\dfrac{1}{x} = 2\;{\tan ^{ - 1}}\dfrac{1}{7} + {\cos ^{ - 1}}\dfrac{3}{5}$ , then

$x =$ ___

0%
$\dfrac {24}{117}$
0%
$\dfrac {7}{3}$
0%
$\dfrac {125}{117}$
0%
None of these

Explanation

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

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

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

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

$=\tan ^{-1}\left(\dfrac{\dfrac{2}{7}}{1-\dfrac{1}{49}}\right)+\theta$

$=\tan ^{-1}\left(\dfrac{7}{24}\right)+\tan ^{-1}\left(\dfrac{4}{3}\right)$

$=\tan ^{-1}\left(\dfrac{\dfrac{7}{24}+\dfrac{4}{3}}{1-\dfrac{7}{24}\times \dfrac{4}{3}}\right)$

$=\tan ^{-1}\left(\dfrac{117}{44}\right)$

$\because \sin ^{-1}\left(\dfrac{1}{x}\right)=\tan ^{-1}\left(\dfrac{117}{44}\right)$

Let

$\theta =\tan ^{-1}\left(\dfrac{117}{44}\right)$

$\therefore \tan \theta=\dfrac{117}{44}$

$\therefore \sin \theta=\dfrac{117}{\sqrt{117^2+44^2}}=\dfrac{117}{125}$

$\therefore \theta =\sin ^{-1}\left(\dfrac{117}{125}\right)$

$\therefore \sin ^{-1}\left(\dfrac{1}{x}\right)=\sin ^{-1}\left(\dfrac{117}{125}\right)$

On comparing

$\dfrac{1}{x}=\dfrac{117}{125}$

$\boxed{\therefore x=\dfrac{125}{117}}$

2116903_1507505_ans_b5b23362f4624b96ac55474eb3fa587c.png

$\cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } =\alpha$ where

$-1-1\le x\le 1,-2\le y\le 2,x\le \cfrac { y }{ 2 }$ then for all

$4{ x }^{ 2 }-4xy\cos { \alpha } +{ y }^{ 2 }$ is equal to

0%
$4\sin ^{ 2 }{ \alpha } -2{ x }^{ 2 }{ y }^{ 2 }$
0%
$4\cos ^{ 2 }{ \alpha } +2{ x }^{ 2 }{ y }^{ 2 }$
0%
$4\sin ^{ 2 }{ \alpha }$
0%
$2\sin ^{ 2 }{ \alpha }$

Explanation

$\cos { \left( \cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } \right) } =\cos { \alpha }$

$\cos { \alpha } \Rightarrow x\times \cfrac { y }{ 2 } +\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\cfrac { { y }^{ 2 } }{ 4 } } \Rightarrow { \left( \cos { \alpha } -\cfrac { xy }{ 2 } \right) }^{ 2 }=\left( 1-{ x }^{ 2 } \right) \left( 1-\cfrac { { y }^{ 2 } }{ 4 } \right)$

${ x }^{ 2 }+\cfrac { { y }^{ 2 } }{ 4 } -xy\cos { \alpha } =1-\cos ^{ 2 }{ \alpha } =\sin ^{ 2 }{ \alpha } \quad$

The value of

$tan \left (\cos^{-1} \dfrac {3}{5} + \tan^{-1} \dfrac {1}{4}\right )$ is

0%
$\dfrac {19}{8}$
0%
$\dfrac {8}{19}$
0%
$\dfrac {19}{12}$
0%
$\dfrac {3}{4}$

Explanation

Given,

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

Let,

${\cos}^{-1}{\dfrac{3}{5}}=a$

$\Rightarrow\,\cos{a}=\dfrac{3}{5}$

$\Rightarrow\,{\sin}^{2}{a}=1-{\cos}^{2}{a}=1-\dfrac{9}{25}=\dfrac{25-9}{25}=\dfrac{16}{25}$

$\therefore\,\sin{a}=\dfrac{4}{5}$

$\Rightarrow\,\tan{a}=\dfrac{\sin{a}}{\cos{a}}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}$

$\Rightarrow\,a={\tan}^{-1}{\dfrac{4}{3}}$

$\therefore\,\tan{\left({\cos}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}$

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

$a={\cos}^{-1}{\dfrac{3}{5}}$

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

$a={\cos}^{-1}{\dfrac{3}{5}}={\tan}^{-1}{\dfrac{4}{3}}$

We know that

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

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

$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{16+3}{12}}{1-\dfrac{4}{12}}\right)}\right)}$

$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{19}{12}}{\dfrac{12-4}{12}}\right)}\right)}$

$=\tan{\left({\tan}^{-1}{\dfrac{19}{8}}\right)}$

$=\dfrac{19}{8}$

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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