MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 3

The number of solutions of the equation

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

0%
$0$
0%
Infinite
0%
$2$
0%
$4$

Explanation

The maximum value of the function is obtained when the value of

$\sqrt {x^2-x+1}$ is equal to

$1$

and value of

$\sqrt {x^2-x}$ is equal to

$0$

$\displaystyle \sin ^{-1}\sqrt{x}, \cos ^{-1}\sqrt{x}$ are defined for

$\displaystyle x\leq 1$ and

$\displaystyle x\geq 0$

$\displaystyle \therefore \sqrt{x^{2}-x+1}\leq 1$ and

$\displaystyle \sqrt{x^{2}-x}\geq 0$

$\displaystyle \Rightarrow x^{2}-x\leq 0$ and

$\displaystyle x^{2}-x\geq 0$

Thus, reducing the equation to

$\displaystyle \Rightarrow x^{2}-x=0$

$\displaystyle \Rightarrow x=1, 0$

$\displaystyle \therefore$ there are two solutions, both satisfies the equation.

The value of

$\displaystyle \tan^{-1} \left( \frac{\sin{ 2} - 1}{\cos {2}} \right )$ is equal to

0%
$\displaystyle \frac{\pi}{2} - 1$
0%
$\displaystyle 2 - \frac{\pi}{2}$
0%
$\displaystyle 1- \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{4} - 1$

Explanation

Given,

$\displaystyle \tan ^{ -1 }{ \left( \frac { \sin { 2 } -1 }{ \cos { 2 } } \right) }$

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

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

$=1-\dfrac { \pi }{ 4 }$

The value of

$\cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$ is

0%
$0$
0%
$8 \pi - 26$
0%
$4 \pi + 2$
0%
None of these

Explanation

The value of

$\cos ^{ -1 }{ \left( \cos { 12 } \right) } -\sin ^{ -1 }{ \left( \sin { 14 } \right) }$

$=\cos ^{ -1 }{ \left( \cos { \left( 4\pi -12 \right) } \right) } -\sin ^{ -1 }{ \left( -\sin { \left( 4\pi-14 \right) } \right) }$ ...................

$(\because \sin(-\theta)=-\sin \theta)$

$=4\pi -12-14+4\pi =8\pi -26$

Find the value of

$x$ if

$\sin (arc \sin x) = \dfrac {\sqrt {2}}{4}$

0%
$\dfrac {\sqrt {2}}{4}$
0%
$\dfrac {\sqrt {7}}{7}$
0%
$\dfrac {\sqrt {2}}{2}$
0%
$\dfrac {2\sqrt {2}}{3}$

Explanation

Given,

$\sin\arcsin { x } =\dfrac { \sqrt { 2 } }{ 4 } =\dfrac { 1 }{ 2\sqrt { 2 } }$

$\therefore arc\sin { x } =arc\sin \left (\dfrac {1}{2\sqrt2}\right)=0.36136$

$\therefore x=\sin(0.36136)=\dfrac { 1 }{ 2\sqrt { 2 } }$

$\displaystyle \tan^{-1}{\left(\frac{a}{x}\right)}+\tan^{-1}{\left(\frac{b}{x}\right)}=\frac{\pi}{2}$ , then

$x$ is equal to :

0%
$ab$
0%
$\dfrac{a}{b}$
0%
$\dfrac{b}{a}$
0%
$\sqrt{ab}$

Explanation

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

$\Rightarrow \tan^{-1}\left(\dfrac{a}{x}\right)=\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{b}{x} \right)=\cot^{-1}\left(\dfrac{b}{x} \right)$

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

$\Rightarrow \tan^{-1}\left(\dfrac{a}{x}\right)=\cot^{-1}\left(\dfrac{b}{x} \right)=\tan^{-1}\left(\dfrac{x}{b}\right)$

$\Rightarrow \dfrac{a}{x}=\dfrac{x}{b}\Rightarrow x^2=ab$

$\Rightarrow x=\pm \sqrt {ab}$

Given

$0 \leq x \leq \frac{1}{2}$ , then the value of

$\tan\left [ \sin^{-1}\left \{ \dfrac{x}{\sqrt{2}}+\dfrac{\sqrt{1-x^2}}{\sqrt{2}} \right \} \right ]$ is.

0%
$\dfrac{\sqrt{1-x^2}+x}{\sqrt{1-x^2}-x}$
0%
$\dfrac{\sqrt{1-x^2}-x}{\sqrt{1-x^2}+x}$
0%
$\dfrac{x+\sqrt{1-x^2}}{x-\sqrt{1-x^2}}$
0%
$\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}$

Explanation

$0\le x\le \dfrac { 1 }{ 2 }$

Let

$\sin ^{ -1 }{ x } =0$

$\Rightarrow 0°\le \theta \le 30°$

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

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

$($ when

$\sin \theta = x$ ,

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

$\tan { \left( \sin ^{ -1 }{ \left( \sin { \theta } .\cos { 45° } +\cos { \theta } .\sin { 45° } \right) } \right) }$

$\tan { \left( \sin ^{ -1 }{ \left( \sin { \left( \theta +45° \right) } \right) } \right) } =\tan { \left( \theta +45° \right) }$

$=\dfrac { \tan { \theta } +\tan { 45° } }{ 1-\tan { \theta } .\tan { 45° } }$

$=\dfrac { \tan { \theta } +1 }{ 1-\tan { \theta } }$

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

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

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

The trigonometric equation

$\sin^{-1}x = 2\sin^{-1}2a$ has a real solution if

0%
$|a| > \dfrac {1}{\sqrt {2}}$
0%
$\dfrac {1}{2\sqrt {2}} < |a| < \dfrac {1}{\sqrt {2}}$
0%
$|a| > \dfrac {1}{2\sqrt {2}}$
0%
$|a| \leq \dfrac {1}{2\sqrt {2}}$

Explanation

$-\pi / 2 \leq \sin^{-1} x\leq \pi / 2 \Rightarrow -\pi /2 \leq 2\sin^{-1} 2a \leq \pi / 2$

$\Rightarrow -\pi / 4 \leq \sin^{-1}2a \leq \pi /4 \Rightarrow -\dfrac {1}{\sqrt {2}} \leq 2a \leq \dfrac {1}{\sqrt {2}}$

$\Rightarrow - \dfrac {1}{2\sqrt {2}} \leq a\leq \dfrac {1}{2\sqrt {2}} \Rightarrow |a|\leq \dfrac {1}{2\sqrt {2}}$

The domain of the function

$f(x)=\sqrt{cos^{-1}\begin{pmatrix}\dfrac{1-|x|}{2}\end{pmatrix}}$ is

0%
$(-3,\,3)$
0%
$[-3,\,3]$
0%
$(-\infty,\,-3)\cup(-3,\,\infty)$
0%
$(-\infty,\,-3)\cup(3,\,\infty)$

Explanation

$f(x)=\sqrt{cos^{-1}\begin{pmatrix}\dfrac{1-|x|}{2}\end{pmatrix}}$

for

$f(x)$ to be defined

$\cos^{-1}{\left(\dfrac{1-|x|}{2}\right)}\ge0$

$\Rightarrow \dfrac{-\pi}{2}\le \cos^{-1}{\left(\dfrac{1-|x|}{2}\right)}\le \dfrac{\pi}{2}$

$\Rightarrow-1 \le \dfrac{1-|x|}{2} \le 1$

$\Rightarrow\;-2\le 1-|x| \le 2$

$\Rightarrow\;-2-1\le -|x| \le 2-1$

$\Rightarrow\;-3 \le -|x| \le 1$ .

$\Rightarrow\;-1\le|x|\le3$

$\Rightarrow |x|\ge-1$ but

$|x|$ is always positive

$\Rightarrow |x|\le3$

$\Rightarrow -3\le x\le3$

$\Rightarrow\;x\,\in[-3,\,3]$

The trigonometric equation

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

0%
$|a| > \dfrac{1}{\sqrt{2}}$
0%
$\dfrac{1}{2\sqrt{2}} < |a| < \dfrac{1}{\sqrt{2}}$
0%
$|a| > \dfrac{1}{2\sqrt{2}}$
0%
$|a| \le \dfrac{1}{2\sqrt{2}}$

Explanation

We have that

$\displaystyle \frac{-\pi}{2} \le \sin^{-1}x\le \frac{\pi}{2}$

$\Rightarrow \dfrac{-\pi}{2}\le 2\, \sin^{-1}2a\le \dfrac{\pi}{2}$

$\Rightarrow \dfrac{-\pi}{4}\le \sin^{-1}2a\le \dfrac{\pi}{4}$

$\Rightarrow \sin\left(\dfrac{-\pi}{4}\right)\le 2a \le \sin\left(\dfrac{\pi}{4}\right)$

$\Rightarrow \dfrac{-1}{\sqrt{2}}\le 2a \le \dfrac{1}{\sqrt{2}}$

$\Rightarrow \dfrac{-1}{2\sqrt{2}}\le a \le \dfrac{1}{2\sqrt{2}}\Rightarrow |a|\le \dfrac{1}{2\sqrt{2}}$

$\theta =\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } -\tan ^{ -1 }{ x } ,1\le x<\infty$ , then the smallest interval in which

$\theta$ lies is

0%
$\cfrac { \pi }{ 2 } \le \theta \le \cfrac { 3\pi }{ 4 }$
0%
$0\le \theta \le \cfrac { \pi }{ 4 }$
0%
$-\cfrac { \pi }{ 4 } \le \theta \le 0$
0%
$\cfrac { \pi }{ 4 } \le \theta \le \cfrac { \pi }{ 2 }$

Explanation

Given,

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

$=$

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

$\tan ^{ -1 }{ x }$ because

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

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

$=$

$\cfrac { \pi }{ 2 }$

$=$

$tan^{-1}(x)$

$\Rightarrow \theta =\cot ^{ -1 }{ x }$
Since,

$1\le x<\infty$ , therefore

$0\le \theta <\cfrac { \pi }{ 4 }$

$\tan \left [3\tan^{-1}\left (\dfrac {1}{5}\right ), -\dfrac {\pi}{4}\right ]$ is equal to

0%
$-\dfrac {13}{46}$
0%
$-\dfrac {11}{46}$
0%
$-\dfrac {7}{46}$
0%
$-\dfrac {4}{23}$
0%
$-\dfrac {9}{46}$

Explanation

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

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

$= \tan \left \{\tan^{-1} \left (\dfrac {2\times \dfrac {1}{5}}{1 - \dfrac {1}{25}}\right ) + \tan^{-1}\left (\dfrac {\dfrac {1}{5} - 1}{1 + \dfrac {1}{5}}\right )\right \}$

$= \tan \left \{\tan^{-1} \left (\dfrac {2}{5}\times \dfrac {25}{24}\right ) + \tan^{-1} \left (-\dfrac {4}{5}\times \dfrac {5}{6}\right )\right \}$

$= \tan \left \{\tan^{-1}\left (\dfrac {5}{12}\right ) - \tan^{-1}\left (\dfrac {2}{3}\right )\right \}$

$= \tan \left (\tan^{-1} \dfrac {\dfrac {5}{12} - \dfrac {2}{3}}{1 + \dfrac {5}{18}}\right )$

$= \tan \left \{\tan^{-1}\left (\dfrac {-3}{12} \times \dfrac {18}{23}\right )\right \} = -\dfrac {9}{46}$

$\tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } =\cfrac { 2\pi }{ 3 }$ , then

$\cot ^{ -1 }{ x } +\cot ^{ -1 }{ y }$ is equal to

0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \sqrt { 3 } }{ 2 }$
0%
$\pi$

Explanation

We have,

$\tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } =\cfrac { 2\pi }{ 3 }$

$\Rightarrow \cfrac { \pi }{ 2 } -\cot ^{ -1 }{ x } +\cfrac { \pi }{ 2 } -\cot ^{ -1 }{ y } =\cfrac { 2\pi }{ 3 }$

$\cot ^{ -1 }{ x } +\cot ^{ -1 }{ y } =\pi -\cfrac { 2\pi }{ 3 } =\cfrac { \pi }{ 3 }$

$2 sin h^{-1}\left ( \dfrac{a}{\sqrt{1-a^{2}}} \right ) = log\left ( \dfrac{1+X}{1-X} \right )$ then X=

0%
a
0%
$\dfrac{1}{a}$
0%
$\sqrt{1-a^{2}}$
0%
$\dfrac{1}{\sqrt{1-a^{2}}}$

Explanation

As we know that

$\sinh^{-1} x=\log(x+\sqrt{x^2+1})$

$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=2\log\bigg(\dfrac{a}{\sqrt{1-a^2}}+\sqrt{\dfrac{a^2}{1-a^2}+1}\bigg)$

$=2\log \bigg(\dfrac{a+\sqrt{a^2+1-a^2}}{\sqrt{1-a^2}}\bigg)$

$=2\log \bigg(\dfrac{a+1}{\sqrt{1-a^2}}\bigg)$

$=\log\bigg(\dfrac{(1+a)^2}{(1-a)(1+a)}\bigg)$

$=\log \bigg(\dfrac{1+a}{1-a}\bigg)$

But given

$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=\log \bigg(\dfrac{1+X}{1-X}\bigg)$

$X=a$

The value of

$x$ satisfying the equation

$\tan ^{ -1 }{ x } +\tan ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) } =\tan ^{ -1 }{ \left( \cfrac { 7 }{ 4 } \right) }$ is equal to

0%
$\cfrac { 1 }{ 2 }$
0%
$-\cfrac { 1 }{ 2 }$
0%
$\cfrac { 3 }{ 2 }$
0%
$-\cfrac { 1 }{ 3 }$
0%
$\cfrac { 1 }{ 3 }$

Explanation

We have

$\quad \tan ^{ -1 }{ x } +\tan ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) } =\tan ^{ -1 }{ \left( \cfrac { 7 }{ 4 } \right) }$

$\tan ^{ -1 }{ \left[ \cfrac { x+\cfrac { 2 }{ 3 } }{ 1-\cfrac { 2 }{ 3 } x } \right] } =\tan ^{ -1 }{ \left( \cfrac { 7 }{ 4 } \right) }$

$\Rightarrow \cfrac { \cfrac { 3x+2 }{ 3 } }{ \cfrac { 3-2x }{ 3 } } =\cfrac { 7 }{ 4 } \quad$

$\Rightarrow 4(3x+2)=7 (3-2x)$

$\Rightarrow x=\cfrac { 13 }{ 26 } =\cfrac { 1 }{ 2 }$

The value of

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

0%
$\dfrac { 3+\sqrt { 5 } }{ 2 }$
0%
$3+\sqrt { 5 }$
0%
$\dfrac { 1 }{ 2 } \left( 3-\sqrt { 5 } \right)$
0%
None of these

Explanation

Let

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

$\cos { \alpha } =\dfrac { \sqrt { 5 } }{ 3 }$ , where

$0 < \alpha < \dfrac { \pi }{ 2 }$ .
Now,

$\tan { \dfrac { \alpha }{ 2 } } =\sqrt { \dfrac { 1-\cos { \alpha } }{ 1+\cos { \alpha } } } =\sqrt { \dfrac { 1-{ \sqrt { 5 } }/{ 3 } }{ 1+{ \sqrt { 5 } }/{ 3 } } }$

$=\sqrt { \dfrac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } } =\sqrt { \dfrac { { \left( 3-\sqrt { 5 } \right) }^{ 2 } }{ 9-5 } }$

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

$ab > -1, bc > -1$ and

$ca > -1$ , then the value of

$\cot^{-1}\left (\dfrac {ab + 1}{a - b}\right ) + \cot^{-1}\left (\dfrac {bc + 1}{b - c}\right ) + \cot^{-1}\left (\dfrac {ca + 1}{c - a}\right )$ is

0%
$-1$
0%
$\cot^{-1}(a + b + c)$
0%
$\cot^{-1}(abc)$
0%
$0$
0%
$\tan^{-1}(a + b + c)$

Explanation

Given,

$ab > -1, bc > -1$ and

$ca > -1$ , then

$\cot^{-1}\left (\dfrac {ab + 1}{a - b}\right ) + \cot^{-1}\left (\dfrac {bc + 1}{b - c}\right ) + \cot^{-1}\left (\dfrac {ca + 1}{c - a}\right )$

$= (\cot^{-1} b - \cot^{-1}a) + (\cot^{-1}c - \cot^{-1} b) + (\cot^{-1} a - \cot^{-1}c)$

$= 0$ .

$\cos^{-1}\left (\dfrac {1 - x^{2}}{1 + x^{2}}\right ) + \cos^{-1}\left (\dfrac {1 - y^{2}}{1 + y^{2}}\right ) = \dfrac {\pi}{2}$ , where

$xy < 1$ , then

0%
$x - y - xy = 1$
0%
$x - y + xy = 1$
0%
$x + y - xy = 1$
0%
$x + y + xy = 1$
0%
$y - x - xy = 1$

Explanation

Given,

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

$\Rightarrow 2\tan^{-1} x + 2\tan^{-1} y = \dfrac {\pi}{2}$

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

$\Rightarrow \dfrac {x + y}{1 - xy} = \tan \dfrac {\pi}{4} = 1$

$\Rightarrow x + y = 1 - xy$

$\Rightarrow x + y + xy = 1$

$\tan^{-1} (-x) + \cos^{-1}\left (\dfrac {-1}{2}\right ) = \dfrac {\pi}{2}$ , them the value of

$x$ is

0%
$\sqrt {3}$
0%
$\dfrac {-1}{\sqrt {3}}$
0%
$\dfrac {1}{\sqrt {3}}$
0%
$-\sqrt {3}$
0%
$1$

Explanation

Given,

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

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

$\Rightarrow -\tan^{-1} x = \dfrac {\pi}{2} - \dfrac {2\pi}{3} = -\dfrac {\pi}{6}$

$\Rightarrow \tan^{-1}x = \dfrac {\pi}{6}$
Therefore,

$x = \tan \dfrac {\pi}{6} = \dfrac {1}{\sqrt {3}}$

$\cos^{-1}\left (\cos \left (\dfrac {7\pi}{5}\right)\right)$ is equal to

0%
$\dfrac{3 \pi}{5}$
0%
$\dfrac{2 \pi}{5}$
0%
$\dfrac{-7 \pi}{5}$
0%
$\dfrac{7 \pi}{5}$
0%
$\dfrac{-2 \pi}{5}$

Explanation

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

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

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

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

$\cos^{-1}\left(\cos \cfrac{3\pi}{5}\right)=\cfrac{3\pi}{5}$

Hence, A is the correct option.

If the non-zero numbers

$x,y,z$ are

$AP$ and

$\tan ^{ -1 }{ x } ,\tan ^{ -1 }{ y } ,\tan ^{ -1 }{ z }$ are also in

$AP$ , then

0%
$xy=yz$
0%
${z}^{2}=xy$
0%
$x=y=z$
0%
${ x }^{ 2 }=yz$

Explanation

We have

$x,y,z$ are in AP

$\Rightarrow 2y=x+z...(i)$
Also,

$\Rightarrow 2\tan ^{ -1 }{ y } =\tan ^{ -1 }{ z } +\tan ^{ -1 }{ x }$

$\Rightarrow \tan ^{ -1 }{ \left( \cfrac { 2y }{ 1-{ y }^{ 2 } } \right) } =\tan ^{ -1 }{ \left( \cfrac { x+z }{ 1-xz } \right) }$

$\Rightarrow \cfrac { 2y }{ 1-{ y }^{ 2 } } =\cfrac { x+z }{ 1-xz }$

$\Rightarrow 1-{ y }^{ 2 }=1-xz$ (Using Eq (i))

${ y }^{ 2 }=xz...(ii)$

$\Rightarrow$

$x,y,z$ are in G.P
Using Eqs. (i) and (ii) we get

$x=y=z$

The value of

$\cos \left( \sin^{-1} \left( \dfrac {2}{3} \right) \right)$ is equal to :

0%
$\dfrac {\sqrt3}{5}$
0%
$\dfrac {5}{3}$
0%
$\dfrac {5}{\sqrt3}$
0%
$\sqrt { \dfrac {5}{3} }$
0%
$\dfrac {\sqrt5}{3}$

Explanation

The value of

$\cos \left( \sin^{-1} \left( \dfrac {2}{3} \right) \right)$ is

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

$= \cos \left( \cos^{-1}\left( \sqrt{1 - \dfrac{4}{9} } \right) \right)$

$= \cos \left( \cos^{-1} \left( \dfrac {\sqrt5}{3} \right) \right)$

$= \dfrac {\sqrt5}{3}$

If two angles of a triangle are

$\tan ^{ -1 }{ (2) }$ and

$\tan ^{ -1 }{ (3) }$ , then the third angle is

0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 2 }$

Explanation

$\textbf{Step 1 :- Formulating the information from the given question. }$

$\text{Suppose the third angle is p.}$

$\text{So from definition we know that the sum of all total three angles of a triangle is}$ $\pi^c$ .

${\text{As 2}} \times {\text{3 > 1 So we can write }} {\tan^{ - 1}}2 + {\tan^{ - 1}}3 = \pi + {\tan^{ - 1}}\left( {\dfrac{{2 + 3}}{{1 - 2 \times 3}}} \right)$

$\mathbf{\left [ \tan^{-1} X + \tan^{-1}Y = \tan^{-1}\dfrac{X+Y}{1-(X\times Y)} \right ]}$

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

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

$\textbf{Step 2 :- Using formula of sum three angles of a triangle.}$

${\text{So p + }}\dfrac{{3\pi }}{4} = \pi$

$\Rightarrow {\text{p = }}\pi {\text{ - }}\dfrac{{3\pi }}{4} = \dfrac{\pi }{4}$

$\textbf{Hence, The correct option is A}$ .

$\sin { \left[ \cot ^{ -1 }{ (x+1) } \right] } =\cos { \left[ \tan ^{ -1 }{ x } \right] }$ , then

$x$ is equal to

0%
$\cfrac { -1 }{ 2 }$
0%
$\cfrac { 1 }{ 2 }$
0%
$0$
0%
$\cfrac { 9 }{ 2 }$

Explanation

Given :

$\sin { \left[ \cot ^{ -1 }{ (x+1) } \right] } =\cos { \left[ \tan ^{ -1 }{ x } \right] }$ (i)

We know that,

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

$\therefore \sin { \left\{ \cot ^{ -1 }{ \left( x+1 \right) } \right\} } =\sin { \left\{ \sin ^{ -1 }{ \cfrac { 1 }{\sqrt{ x^{ 2 }+2x+2}}} \right\} } =\cfrac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } }$
Also,

$\cos { \left( \tan ^{ -1 }{ x } \right) } =\cos { \left\{ \cos ^{ -1 }{ \cfrac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } \right\} } =\cfrac { 1 }{ \sqrt { 1+{ x }^{ 2 } } }$
Thus from (i), we get

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

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

$\Rightarrow x=-\cfrac { 1 }{ 2 }$

The sum of

$\cot ^{ -1 }{ 2 } +\cot ^{ -1 }{ 8 } +\cot ^{ -1 }{ 18 } ....\infty =\cfrac { \pi }{ \lambda }$ , then

$\lambda$ is

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$\cot^{-1}(2)+\cot^{-1}(8)+\cot^{-1}(18).......\infty =\dfrac{\pi}{\lambda}$

from L.H.S.

$n^{th}$ term is

$\Rightarrow \cot^{-1}(2n^2)$

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

$=\tan^{-1}\left(\dfrac{2}{4n^2}\right)$

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

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

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

Now,

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

$A=2n+1$ ;

$B=2n-1$

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

$n=1, 2, 3$ ...

$\Rightarrow \tan^{-1}(3)-\tan^{-1}(1)+\tan^{-1}(5)-\tan^{-1}(3)+.....\tan^{-1}\infty$

$\Rightarrow \tan^{-1}(\infty)-\tan^{-1}(1)$

$\Rightarrow \dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$

Comparing RHS to

$\lambda =4$ .

1157300_699524_ans_3656c4c8980b4db78334b9930bf0c222.jpg

The domain of function

$f\left( x \right) =\sin ^{ -1 }{ 5x }$ is

0%
$\left( -\dfrac { 1 }{ 5 } ,\dfrac { 1 }{ 5 } \right)$
0%
$\left[ -\dfrac { 1 }{ 5 } ,\dfrac { 1 }{ 5 } \right]$
0%
$R$
0%
$\left( 0,\dfrac { 1 }{ 5 } \right)$

Explanation

Given,

$f(x)=\sin ^{-1}5x$

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

$\therefore -1\le 5x\le 1$

$\Rightarrow \dfrac { -1 }{ 5 } \le x\le \dfrac { 1 }{ 5 }$

Hence, domain is

$\left[ -\dfrac { 1 }{ 5 } ,\dfrac { 1 }{ 5 } \right]$ .

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

0%
$\dfrac{\pi}{4} - \dfrac{1}{2} \tan^{-1} x$
0%
$\dfrac{1}{2} \tan^{-1} x$
0%
$\dfrac{\pi}{2} - \dfrac{1}{2} \tan^{-1} x$
0%
$\dfrac{\pi}{4} + \dfrac{1}{2} \tan^{-1} x$

Explanation

Let,

$x= \tan \theta$

$\therefore 1+x^2=\space 1+\tan^2 \theta = \sec^2 \theta$ .....(i)

Given,

$\tan^{-1} (x+ \sqrt{1+x^2})$

$= \tan^{-1} (\tan \theta +\sqrt {\sec ^2 \theta})$ ....{From (i)}

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

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

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

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

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

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

Dividing by

$\cos \dfrac{\theta}{2}$ on both numerator and denominator, we get

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

$= \tan ^{-1} \tan \left (\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)$

$=\dfrac {\pi}{4}+\dfrac {\theta}{2}$

$= \dfrac{\pi}{4}+\dfrac{1}{2}\tan^{-1}x$ ....Resubstituting value of

$\theta$

If $2\sinh ^{ -1 }{ \left( \dfrac { a }{ \sqrt { 1-{ a }^{ 2 } } } \right) } =\log { \left( \dfrac { 1+x }{ 1-x } \right) }$ , then $x=$

0%
$a$
0%
$\dfrac { 1 }{ a }$
0%
$\sqrt { 1-{ a }^{ 2 } }$
0%
$\dfrac { 1 }{ \sqrt { 1-{ a }^{ 2 } } }$

Explanation

As we know that

$\sinh^{-1} x=\log(x+\sqrt{x^2+1})$

$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=2\log\bigg(\dfrac{a}{\sqrt{1-a^2}}+\sqrt{\dfrac{a^2}{1-a^2}+1}\bigg)$

$=2\log \bigg(\dfrac{a+\sqrt{a^2+1-a^2}}{\sqrt{1-a^2}}\bigg)$

$=2\log \bigg(\dfrac{a+1}{\sqrt{1-a^2}}\bigg)$

$=\log\bigg(\dfrac{(1+a)^2}{(1-a)(1+a)}\bigg)$

$=\log \bigg(\dfrac{1+a}{1-a}\bigg)$

But given

$2\sinh^{-1} \bigg(\dfrac{a}{\sqrt{1-a^2}}\bigg)=\log \bigg(\dfrac{1+x}{1-x}\bigg)$

$x=a$

The value of

$\sec ^{ 2 }{ \left( \tan ^{ -1 }{ 2 } \right) } +\text{cosec} ^{ 2 }{ \left( \cot ^{ -1 }{ 3 } \right) }=$ ____

0%
$6$
0%
$15$
0%
$13$
0%
$25$

Explanation

Let

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

$2=\tan x$

Let

$perpendicular=p=2k$

$base=b=k$

$hypotenuse=h=\sqrt{(2k)^2+{k}^2}=\sqrt{(5k)^2}=\sqrt{5} k$

$\sec x=\dfrac{hypotenuse}{base}=\dfrac{\sqrt{5}k}{k}=\sqrt{5}$

$x=sec^{-1}\sqrt{5}$ (2)

Let

$\cot^{-1}3=y$ (3)

$3=\cot y$

Let

$base=3=k^{'}$

$perpendicular=p=k^{'}$

$hypotenuse=h=\sqrt{(k^{'})^2+{(3k^{'}})^2}=\sqrt{(10k^{'})^2}=\sqrt{10} k^{'}$

$\text{cosec} y=\dfrac{hypotenuse}{perpendicular}=\dfrac{\sqrt{10}k^{'}}{k^{'}}=\sqrt{10}$

$y=\text{cosec}^{-1}\sqrt{10}$ (4)

From (1),(2),(3) and (4)

$\sec^{2}(\tan^{-1}2)+\text{cosec}^{2}(\cot^{-1}3)=\sec^{2}x+\text{cosec}^{2}y=\sec^{2}sec^{-1}\sqrt{5}+\text{cosec}^{2}\text{cosec}^{-1}\sqrt{10}$

$=(\sec \sec^{-1}\sqrt{5})^2+(\text{cosec} \text{cosec}^{-1}\sqrt{10})^{2}=(\sqrt{5})^{2}+(\sqrt{10})^{2}=5+10=15$

The value of $\cot^{-1} \left[ \dfrac{\sqrt{1 - \sin x} +\sqrt{1 + \sin x}}{\sqrt{(1 - \sin x)} - \sqrt{(1 + \sin x)}} \right]$ is

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

Explanation

We have

$\cot^{-1} \, \left[ \dfrac{\sqrt{1 - \sin x} +\sqrt{1 + \sin x}}{\sqrt{(1 - \sin x)} - \sqrt{(1 + \sin x)}} \right]$

$= \cot^{-1} \left[ \dfrac{ \sqrt {\left( \cos \dfrac{x}{2} - \sin \dfrac{x}{2} \right)^2} + \sqrt {\left( \cos \dfrac{x}{2} + \sin \dfrac{x}{2} \right)^2} }{ \sqrt {\left( \cos \dfrac{x}{2} - \sin \dfrac{x}{2} \right)^2} - \sqrt {\left( \cos \dfrac{x}{2} + \sin \dfrac{x}{2} \right)^2 }} \right]$

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

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

$\sin^{-1} \sqrt{2 - x} = \cos^{-1} \sqrt{x-1}$ holds for all real x.

0%
True
0%
False

Explanation

The equality will hold only for those values of x for which

$0 \leq 2 - x \leq 1$ and

$0 \leq x - 1 \leq 1$ ,
that is, for

$1 \leq x \leq 2$ and not for all real values of x.

The value of

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

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

Explanation

$E = \pi - cos^{-1} (1) - sin^{-1} (1) = \pi - \dfrac{\pi}{2} = \dfrac{\pi}{2}$

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

and

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

Range of the function

$f(x)=4{ tan }^{ -1 }x+3{ sin }^{ -1 }x+{ sec }^{ -1 }x$ is

0%
$\left\{ \frac { -3\pi }{ 2 } ,\frac { -5\pi }{ 2 } \right\}$
0%
$\left\{ \frac { 3\pi }{ 2 } ,\frac { 5\pi }{ 2 } \right\}$
0%
$\left\{ \frac { 3\pi }{ 2 } ,\frac { -5\pi }{ 2 } \right\}$
0%
$\left\{ \frac { -3\pi }{ 2 } ,\frac { 5\pi }{ 2 } \right\}$

$y= \sec(\tan^{-1} x)$ then,

$\dfrac{dy}{dx}$ at

$x=1$ is equal to

0%
$\dfrac{1}{2}$
0%
$1$
0%
$\sqrt2$
0%
$\frac{1}{\sqrt2}$

Explanation

$\\ y=sec(tan^{-1}x)\\\therefore sec(tan^{-1}x)tan(tan^{-1}x)\cdot (\frac{1}{1+x^2})\\ put x=1\\ then(\frac{dy}{dx})=sec(tan^{-1}1)tan(tan^{-1}1)\cdot (\frac{1}{1+1^2})\\=sec(\frac{\pi}{4})\cdot tan(\frac{\pi}{4})\cdot (\frac{1}{2})\\= \sqrt 2\cdot 1 \cdot (\frac{1}{2})=(\frac{1}{\sqrt2})$

Number of solution of the equation

$\cos^{-1}(1-x)-2\cos^{-1}x=\dfrac{\pi}{2}$ is

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

${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = {\pi \over 2},x \in \mathop R\limits^ \bullet$

0%
True
0%
False

Explanation

Given

$\tan ^{-1} +\cot ^ {-1 }x$

Let

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

$\implies \tan A =x$

$\implies \cot \left(\dfrac \pi 2 -A\right) =x$

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

$\implies \tan ^ {-1} x+\cot ^ {-1} x$

$\implies A+\dfrac {\pi }2 -A$

$\implies \dfrac \pi 2$

The value of

$k$ if the equation

$kx + \ \sin^{-1}(x^2-8x+17) + \ \cos^{-1}(x^2-8x+17) = \frac{9\pi}{2}$ has atleast one solution is

0%
$2\pi$
0%
$\pi$
0%
$1$
0%
$\frac{\pi}{2}$

Explanation

$\implies x^{2}-8{x}+17=(x-4)^{2}+1\in [-1,1]$

$\implies (x-4)^{2}\in [-2,0]\implies x-4=0\implies x=4$

$k{x}+\text{sin}^{-1}(x^{2}-8{x}+17)+\text{cos}^{-1}(x^{2}-8{x}+17)=\dfrac{9\pi}{2}$

$4{k}+\dfrac{\pi}{2}=\dfrac{9\pi}{2}\implies 4{k}=4\pi\implies k=\pi$

The value of

$\cot \left( \dfrac{\pi}{4} - 2 \cot^{-1} 3\right)$ , is:

0%
$1$
0%
$7$
0%
$-1$
0%
None of these

Explanation

Let

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

$\Rightarrow \cot x=3$

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

$=\dfrac {1}{\tan \left (\dfrac {\pi} {4}-2x\right)} =\dfrac {1+\tan2x}{1-\tan 2x}$

Now,

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

Final answer

$=\dfrac {1+\dfrac {3}{4}}{1-\dfrac {3}{4}}=7$

$\tan^{-1}(\dfrac{1}{4}) + \tan^{-1}(\dfrac{2}{9})$ is equal to

0%
$\dfrac{1}{2} \ \cos^{-1}(\dfrac{3}{5})$
0%
$\dfrac{1}{2} \sin^{-1}(\dfrac{4}{5})$
0%
$\tan^{-1}(\dfrac{1}{2})$
0%
$\cos^{-1}(\dfrac{8}{9})$

Explanation

$\tan^{-1}\left(\cfrac{1}{4}\right)+\tan^{-1}\left(\cfrac{2}{9}\right)$

$=\tan^{-1}\left(\cfrac{\cfrac{1}{4}+\cfrac{2}{9}}{1-\cfrac{1}{4}\times\cfrac{2}{9}}\right)$

$=\tan^{-1}\left(\cfrac{\cfrac{1}{4}+\cfrac{2}{9}}{\cfrac{36-2}{36}}\right)$

$=\tan^{-1}\left(\cfrac{\cfrac{9+8}{36}}{\cfrac{34}{36}}\right)$

$=\tan^{-1}\left(\cfrac{1}{2}\right)$ or

$\sin^{-1}\cfrac{1}{\sqrt5}$ or

$\cos^{-1}\cfrac{2}{\sqrt5}$

$\sin^{-1} x + \sin ^{-1} y + \sin ^{-1} z = \pi$ , then the value of

$x\sqrt{1-x^2}+ y\sqrt{1-y^2}+ z\sqrt{1- z^2}$ will be:

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

$\sum^n_{m = 1} \tan^{-1} \left(\dfrac{2m}{m^4 + m^2 + 2} \right)$ is equal to

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

Explanation

$\displaystyle \sum^n_{m = 1} \tan^{-1} \left(\dfrac{2m}{m^4 + m^2 + 2} \right)$ is equal to

In General formula for

$tan^{-1}x = tan^{-1}y$ is

$tan^{-1}\left ( \dfrac{x\pm y}{1\pm xy} \right )...(1)$

Now, given summation is,

$\displaystyle \sum_{m = 1}^{n} tan^{-1}\left(\dfrac{2m}{m^{4}+m^{2}+2}\right) = \sum_{m = 1}^{n} tan^{-1}\left(\dfrac{2m}{1+(m^{4}+m^{2}+1)}\right)$

Now, factorisation of.

$m^{4}+m^{2}+1$ is

$m^{4}+m^{2}+1 = (m^{2}+m+1) \times (m^{2}-m+1)$

and difference between

$(m^{2}-m+1)$ and

$(m^{2}-m+1)$ is 2m.

$\displaystyle \Rightarrow \sum_{m=1}^{n} tan^{-1}\left ( \dfrac{(m^{2}+m+1)-(m^{2}-m+1)}{1+(m^{2}+m+1)(m^{2}-m+1)} \right )...(2)$

Now, compare

$eq^{n} (1)$ &

$eq^{n}(2)$

$\displaystyle = \sum_{m-1}^{n} tan^{-1}(m^{2}+m+1)-tan^{-1}(m^{2}*m+1)$

Now putting value from 1, 2, ...,n.

$= tan^{-1}(m^{2}+1+1)-tan^{-1}(1^{2}-1+1)+tan^{-1}(2^{2}+2+1)$

$-tan^{-1}(2^{2}-2+1)+...+tan^{2-1}(n^{2}+n+1)..tan^{-1}(n^{2}-n+1)$

$= tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(7)-tan^{-1}(3)+...+tan^{-1}(n^{2}+n+1)-tan^{-1}(n^{2}-n+1)$

here alternative terms will canceled and we are left with

$tan^{-1}(n^{2}+n+1)-tan^{-1}(1)$

So,

$\boxed{tan^{-1}(n^{2}+n+1)-\dfrac{\pi}{4}}$

1091872_1035489_ans_5d525eb1fd684fe390676dcf813605c7.jpg

$a sin^{-1} x -b cos^{-1}x =c$ , then the value of

$a sin ^{-1} x + b cos^{-1} x$ (whenever exists) is equal to

0%
$0$
0%
$\frac{\pi ab + c(b-a)}{a+b}$
0%
$\frac{\pi}{2}$
0%
$\frac{\pi ab + c(a-b)}{a+b}$

Explanation

As we know that

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

Given

$a\sin^{-1} x-b\cos^{-1} x=c$

$\implies a\sin^{-1} x-b(\frac{\pi}{2}-\sin^{-1} x)=c$

$\implies (a+b)\sin^{-1} x=c+\dfrac{b\pi}{2}$

$\implies \sin^{-1} x=\dfrac{2 c+b\pi}{2(a+b)}$

$a\sin^{-1} x+b\cos^{-1} x=a\sin^{-1} x+b(\frac{\pi}{2}-\sin^{-1} x)$

$=(a-b)\sin^{-1} x+b\dfrac{\pi}{2}$

$=\dfrac{(a-b)(2 c+b\pi)}{2(a+b)}+\dfrac{b\pi}{2}$

$=\dfrac{2 c(a-b)+b\pi(a-b+a+b)}{2(a+b)}$

$=\dfrac{c(a-b)+a b \pi}{(a+b)}$

$\tan^{-1}\dfrac{a+x}{a}+\tan^{-1}\dfrac{a-x}{a}=\dfrac{\pi}{6}$ , then

$x^2=$ ?

0%
$2\sqrt{3}a$
0%
$\sqrt{3}a$
0%
$2\sqrt{3}a^2$
0%
None of these

Explanation

$\tan^{-1}\left(\dfrac{a+x}{a}\right)+\tan^{-1}\left(\dfrac{a-x}{a}\right)=\dfrac{\pi }{6}$

$\tan^{-1}\dfrac{\left(\dfrac{a+x}{a}+\dfrac{a-x}{a}\right)}{\left(1-\left(\dfrac{a+x}{a}\right)\left(\dfrac{a-x}{a}\right)\right)}=\dfrac{\pi }{6}$

$\tan^{-1}\left(\dfrac{\dfrac{2a}{a}}{1\dfrac{1a^{2}-x^{2}}{a^{2}}}\right)=\dfrac{\pi }{6}$

$\tan^{-1}\left(\dfrac{2a^{2}}{a^{2}-a^{2}+x^{2}}\right)=\dfrac{\pi }{6}$

$\dfrac{2a^{2}}{x^{2}}=\tan\dfrac{\pi }{6}$

$x^{2}=\dfrac{2a^{2}}{\dfrac{1}{\sqrt{3}}}$

$x^{2}=2\sqrt{3}a^{2}$

The value of

$\displaystyle {\sin ^{ - 1}}\left( {\frac{{12}}{{13}}} \right)+ {\cos ^{ - 1}}\left( {\frac{4}{5}} \right) + {\tan ^{ - 1}}\left( {\frac{{63}}{{16}}} \right)$

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

Explanation

$\sin^{-1}\left[\cfrac{12}{13}\right]+\cos^{-1}\left[\cfrac{4}{5}\right]+\tan^{-1}\left[\cfrac{63}{16}\right]$

Let

$\sin^{-1}\left(\cfrac{12}{13}\right)=\theta$

$\sin\theta=\left(\cfrac{12}{13}\right)$

$\cos\theta=\sqrt{1-\left(\cfrac{12}{13}\right)^{2}}=\sqrt{\cfrac{169-144}{169}}=\cfrac{5}{13}$

$\tan\theta=\cfrac{12}{5}$

Let

$\cos\beta=\cfrac{4}{5}$

$\sin\beta=\sqrt{1-\cfrac{16}{25}}=\cfrac{3}{5}$

$\tan\beta=\cfrac{\cfrac{3}{5}}{\cfrac{4}{5}}=\cfrac{3}{4}$

Now,

$\tan^{-1}\cfrac{12}{5}+\tan^{-1}\cfrac{3}{4}+\tan^{-1}\cfrac{63}{16}$

$\tan^{-1}\left(\cfrac{\cfrac{12}{5}+\cfrac{3}{4}}{1-\cfrac{12}{5}\times \cfrac{3}{4}}\right)+\tan^{-1}\cfrac{63}{16}$

$\tan^{-1}\left[\cfrac{48+15}{20-36}\right]+\tan^{-1}\left(\cfrac{63}{16}\right)$

$-\tan^{-1}\cfrac{63}{16}+\tan^{-1}\cfrac{63}{16}=0=\tan\pi$ .

Solve:

${\sin ^{ - 1}}\dfrac{4}{5} + {\sin ^{ - 1}}\dfrac{5}{{13}} + {\sin ^{-1}}\dfrac{{16}}{{65}}$

0%
$\dfrac {\pi}{2}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{8}$

Explanation

${\sin ^{ - 1}}\left( {{4 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 \over {13}}} \right) + {\sin ^{ - 1}}\left( {{{16} \over {65}}} \right) = {\pi \over 2}$

${\sin ^{ - 1}}A + {\sin ^{ - 1}}B = {\sin ^{ - 1}}\left( {A\sqrt {1 - {B^{.2}}} + B\sqrt {1 - {B^2}} } \right)$

${\sin ^{ - 1}}\left( {{4 \over 5}\sqrt {1 - {{\left( {{5 \over {12}}} \right)}^2}} + {5 \over {13}}\sqrt {1 - {{\left( {{4 \over 5}} \right)}^2}} } \right)$

${\sin ^{ - 1}}\left( {{{63} \over {65}}} \right)$

${\sin ^{ - 1}}\left( {{{63} \over {65}}} \right) + {\sin ^{ - 1}}{{16} \over {65}}$

Applying some formula

${\sin ^{ - 1}}\left( {{{63} \over {65}}\sqrt {1 - {{\left( { - {{16} \over {65}}} \right)}^2}} + {{16} \over {65}}\sqrt {1 - {{\left( {{{63} \over {65}}} \right)}^2}} } \right)$

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

${\sin ^{ - 1}}\left( {\sin \pi /2} \right)$

$\pi /2$

$\sin ^{ -1 }{ \left( x-\displaystyle\frac { { x }^{ 2 } }{ 2 } +\displaystyle\frac { { x }^{ 3 } }{ 4 } -....\infty \right) } +\cos ^{ -1 }{ \left( { x }^{ 2 }-\displaystyle\frac { { x }^{ 4 } }{ 2 } +\displaystyle\frac { { x }^{ 6 } }{ 4 } -....\infty \right) } =\displaystyle\frac { \pi }{ 2 }$ and

$0<x<\sqrt { 2 }$ then

$x$ =

0%
$\displaystyle\frac { 1 }{ 2 }$
0%
$1$
0%
$\displaystyle\frac{- 1 }{ 2 }$
0%
$-1$

Explanation

$x-\displaystyle\frac { { x }^{ 2 } }{ 2 } +\displaystyle\frac { { x }^{ 3 } }{ 4 } -....\infty$ =

${ x }^{ 2 }-\displaystyle\frac { { x }^{ 4 } }{ 2 } +\displaystyle\frac { { x }^{ 6 } }{ 4 } -....\infty$

$\quad x={ x }^{ 2 }\Rightarrow x=1$

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

$x$ =

0%
${{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{5}}{3\sqrt{50}} \right\}$
0%
${{\cos }^{-1}}\left\{ \dfrac{10-\sqrt{5}}{3\sqrt{50}} \right\}$
0%
${{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{15}}{3\sqrt{50}} \right\}$
0%
None of these

Explanation

We have,

${{\cos }^{-1}}\dfrac{2}{3}+{{\tan }^{-1}}\dfrac{1}{7}$

We know that,

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

Therefore,

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

$={{\cos }^{-1}}\dfrac{2}{3}+{{\cos }^{-1}}\left( \dfrac{7}{\sqrt{50}} \right)$

We know that,

${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left\{ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right\}$

Therefore,

$={{\cos }^{-1}}\left\{ \dfrac{2}{3}\times \dfrac{7}{\sqrt{50}}-\sqrt{1-{{\left( \dfrac{2}{3} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{7}{\sqrt{50}} \right)}^{2}}} \right\}$

$={{\cos }^{-1}}\left\{ \dfrac{14}{3\sqrt{50}}-\sqrt{\dfrac{5}{9}}\sqrt{\dfrac{1}{50}} \right\}$

$={{\cos }^{-1}}\left\{ \dfrac{14}{3\sqrt{50}}-\dfrac{\sqrt{5}}{3\sqrt{50}} \right\}$

$={{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{5}}{3\sqrt{50}} \right\}$

Hence, this is the answer.

$\sin ^{ -1 }{ x } =\cfrac { \pi }{ 6 }$ then find

$'x'$

0%
$\dfrac {\sqrt 3}{2}$
0%
$\dfrac {1}{2}$
0%
$1$
0%
None of these

Explanation

$\textbf{Find the value of x}$

$\text{We have the given data,}$

$\sin^{-1} x=\dfrac{\pi}{6}$

$\text{Taking sine on both sides we get,}$

$\sin(\sin^{-1 } x)=\sin\dfrac{\pi}{6}$

$\Rightarrow x=\sin\dfrac{\pi}{6}$

$\Rightarrow x=\dfrac{1}{2}$

$\textbf{Hence value of x is}$

$\boldsymbol{\dfrac{1}{2}}$

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

$\tan^{ - 1}x - \tan^{ - 1}y = 0$ then value of

${x^2} + ax + {y^2}$ is where a=

$\dfrac{1}{\sqrt{2}}$

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

Explanation

$\cos ^{ -1 }{ x } +\cos ^{ -1 }{ y } =\cfrac { \pi }{ 2 } \Rightarrow 2\cos ^{ -1 }{ x } =\cfrac { \pi }{ 2 } \Rightarrow \cos ^{ -1 }{ x } =\cfrac { \pi }{ 4 } \quad$

$\tan ^{ -1 }{ x } -\tan ^{ -1 }{ y } =0\Rightarrow \tan ^{ -1 }{ x } =\tan ^{ -1 }{ y } \Rightarrow x=y\quad$

$x=\cfrac { 1 }{ \sqrt { 2 } } ;y=\cfrac { 1 }{ \sqrt { 2 } }$

${ x }^{ 2 }+ax+{ y }^{ 2 }=\cfrac { 1 }{ 2 } +\cfrac { 1 }{ \sqrt { 2 } } \times \cfrac { 1 }{ \sqrt { 2 } } +\cfrac { 1 }{ 2 }$

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {\dfrac{{2r}}{{1 - {r^2} + {r^4}}}} \right)}$ is equal to

0%
$\dfrac{\pi} {4}$
0%
$\dfrac{\pi} {2}$
0%
$\dfrac{{3\pi }}{4}$
0%
none of these

Explanation

$\lim {n\to \infty } \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ \left(\dfrac { 2r }{ 1-{ r }^{ 2 }+{ r }^{ 4 } } \right)} }$

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

$=\lim {n\to \infty } \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ \left(\dfrac { r(r+1)-r(r-1) }{ 1+[r({ r }-1)][r(r+1)] } \right) } }$

$=\lim {n\to \infty } \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ (r(r+1)- } } \tan ^{ -1 }{ (r(r-1)) }$

$=\lim {n\to \infty } [tan ^{-1} (1(1+1)- \tan ^{-1} (1(1-1))+tan^{-1} (2(2+1))-\tan ^{-1} (2(2-1))\\ \quad +.......+\tan^{-1}(n(n+1))-\tan^{-1} (n(n-1))]$

$=\lim {n\to \infty } [tan ^{-1} (2)- \tan ^{-1} (0)+tan^{-1} (6))-\tan ^{-1} (2)+.........+\tan^{-1}(n(n+1))-\tan^{-1} (n(n-1))]$

$=\lim {n\to \infty } [- \tan ^{-1} (0)+\tan^{-1}(n(n+1))]$

$=\lim {n\to \infty } [\tan^{-1}(n(n+1))]$

$=\lim {n\to \infty } [\tan^{-1}(n^2+n))]$

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

$=\frac {\pi}{2}$

$\tan \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{5}{7} \right) \right] + \cot \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{5}{7}\right) \right]$ is equal to

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$\dfrac{5}{7}$
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$\dfrac{10}{7}$
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$\dfrac{14}{5}$
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$\dfrac{7}{5}$

Explanation

$\tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\cot \left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]$

$=\tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\tan \left [ \dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]$

We are using the formula:

$\tan \left [ \dfrac{\pi}{4}+\dfrac{x}{2} \right ]+\tan \left [ \dfrac{\pi}{4}-\dfrac{x}{2} \right ]=\dfrac{2}{\cos x}, x=\cos^{-1}\dfrac{a}{b}$

$\tan \left [ \dfrac{\pi}{4}+\dfrac{x}{2} \right ]+\tan \left [ \dfrac{\pi}{4}-\dfrac{x}{2} \right ]=\dfrac{2b}{a}$

$\therefore \tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\cot \left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ] =\dfrac{2\times 7}{5}$

$\Rightarrow \therefore \tan\left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ]+\cot \left [ \dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\left ( \dfrac{5}{7} \right ) \right ] = \dfrac{14}{5}$

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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