If $$tan^{-1}x+2 cot^{-1}x=\dfrac{2\pi}{3}$$, then x is equal to

$$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

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

$$y=-\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$$

$$x=\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 6

The value of

$\tan^{-1}\dfrac{1}{3}+\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{8}$ is ___________.

0%
$\dfrac{11\pi}{5}$
0%
$\dfrac{\pi}{4}$
0%
$\pi$
0%
$\dfrac{3\pi}{4}$

Explanation

$\tan^{-1} \left (\dfrac{1}{3} \right ) + \tan^{-1} \left (\dfrac{1}{5} \right ) + \tan^{-1} \left (\dfrac{1}{7} \right ) + \tan^{-1} \left (\dfrac{1}{8} \right )$

$= \tan^{-1} \left (\dfrac{\dfrac{1}{3} + \dfrac{1}{5}}{1 - \dfrac{1}{15}} \right ) + \tan^{-1} \left (\dfrac{\dfrac{1}{7} + \dfrac{1}{8}}{1 - \dfrac{1}{56}} \right )$

$= \tan^{-1} \left (\dfrac{\dfrac{8}{15}}{\dfrac{14}{15}} \right ) + \tan^{-1} \left (\dfrac{\dfrac{15}{56}}{\dfrac{55}{56}} \right )$

$= \tan^{-1} \left (\dfrac{8}{14} \right ) + \tan^{-1} \left (\dfrac{15}{55} \right )$

$= \tan^{-1} \left (\dfrac{4}{7} \right ) + \tan^{-1} \left (\dfrac{3}{11} \right )$

$= \tan^{-1} \left (\dfrac{\dfrac{4}{7} + \dfrac{3}{11}}{1 - \dfrac{12}{77}} \right )$

$= \tan^{-1} \left (\dfrac{\dfrac{65}{77}}{\dfrac{65}{77}} \right )$

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

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

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

$\tan \left [2\tan^{-1}\dfrac {1}{5}-\dfrac {\pi}{4}\right]=$ ?

0%
$\dfrac {7}{17}$
0%
$\dfrac {-7}{17}$
0%
$\dfrac {7}{12}$
0%
$\dfrac {-7}{12}$

Explanation

Let

$2\tan^{-1}\dfrac {1}{5}=\theta$ .

Then,

$\tan^{-1}\dfrac {1}{5}=\dfrac {1}{2}\theta \Rightarrow \tan \dfrac {\theta}{2} =\dfrac {1}{5}$

$\therefore \ \tan \theta =\dfrac {2\tan \dfrac {1}{2}\theta}{1-\tan^2 \dfrac {1}{2}\theta}=\dfrac {\left(2\times \dfrac {1}{5}\right)}{\left(1-\dfrac {1}{25}\right)}=\left (\dfrac {2}{5}\times \dfrac {25}{24}\right)=\dfrac {5}{12}$

$\therefore \ \tan \left [2\tan^{-1}\dfrac {1}{5}-\dfrac {\pi}{4}\right]=\tan \left (\theta -\dfrac {\pi}{4}\right)=\dfrac {\tan \theta -\tan \dfrac {\pi}{4}}{1+\tan \theta \cdot \tan \dfrac {\pi}{4}}=\dfrac {\left (\dfrac {5}{12}-1\right)}{\left(1+\dfrac {5}{12}\times 1\right)}=\dfrac {\left (\dfrac {-7}{12}\right)}{\left (\dfrac {17}{12}\right)}=\dfrac {-7}{17}$

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

0%
$\dfrac {3}{4}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {3}{5}$
0%
$none\ of\ these$

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

0%
$\dfrac {3}{5}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {4}{9}$
0%
$none\ of\ these$

Explanation

Let

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

$x\in \left(\dfrac {-\pi}{2}, \dfrac {\pi}{2}\right)$ .

$\therefore \ \tan x =\dfrac {3}{4}$ and since

$x \in \left(\dfrac {-\pi}{2}, \dfrac {\pi}{2}\right)$ , where have

$\cos x > 0$

$\therefore \ \cos x =\dfrac {1}{\sec x}=\dfrac {1}{\sqrt {1+\tan^2 x}}=\dfrac {1}{\sqrt {1+\dfrac {9}{16}}}=\dfrac {4}{5}$

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

Evaluate :

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

0%
$\dfrac {(3-\sqrt 5)}{2}$
0%
$\dfrac {(3+\sqrt 5)}{2}$
0%
$\dfrac {(5-\sqrt 3)}{2}$
0%
$\dfrac {(5+\sqrt 3)}{2}$

$x\neq 0$ then

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

0%
$-1$
0%
$1$
0%
$0$
0%
$none\ of\ these$

The value of

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

0%
$0$
0%
$1$
0%
$-1$
0%
$none\ of\ these$

Explanation

As we know that,

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

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

The value of

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

0%
$\dfrac {2}{5}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {-2}{5}$
0%
$none\ of\ these$

$\sin \left [2\tan^{-1}\dfrac {5}{8}\right]$

0%
$\dfrac {25}{64}$
0%
$\dfrac {80}{89}$
0%
$\dfrac {75}{128}$
0%
$none\ of\ these$

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

0%
$\dfrac {12}{25}$
0%
$\dfrac {16}{25}$
0%
$\dfrac {24}{25}$
0%
$none\ of\ these$

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

0%
$1$
0%
$0$
0%
$\dfrac {-1}{2}$
0%
$none\ of\ these$

$\tan \left\{\cos^{-1}\dfrac {4}{5}+\tan^{-1}\dfrac {2}{3}\right\}=$ ?

0%
$\dfrac {13}{6}$
0%
$\dfrac {17}{6}$
0%
$\dfrac {19}{6}$
0%
$\dfrac {23}{6}$

Explanation

$\cos ^{ -1 }{ x } =\tan ^{ -1 }{ \dfrac { \sqrt { 1-{ x }^{ 2 } } }{ x } } \Rightarrow \cos ^{ -1 }{ \dfrac { 4 }{ 5 } } =\tan ^{ -1 }{ \dfrac { \sqrt { 1-\dfrac { 16 }{ 25 } } }{ \left( \dfrac { 4 }{ 5 } \right) } } =\tan ^{ -1 }{ \dfrac { 3 }{ 4 } }$

$\therefore \cos ^{ -1 }{ \dfrac { 4 }{ 5 } } +\tan ^{ -1 }{ \dfrac { 2 }{ 3 } } =\tan ^{ -1 }{ \dfrac { 3 }{ 4 } } +\tan ^{ -1 }{ \dfrac { 2 }{ 3 } } =\tan ^{ -1 }{ \left( \dfrac { \dfrac { 3 }{ 4 } +\dfrac { 2 }{ 3 } }{ 1-\dfrac { 3 }{ 4 } \times \dfrac { 2 }{ 3 } } \right) } =\tan ^{ -1 }{ \dfrac { 17 }{ 6 } }$

$\therefore$ given exp.

$=\tan { \left\{ \tan ^{ -1 }{ \dfrac { 17 }{ 6 } } \right\} } =\dfrac { 17 }{ 6 }$

Evaluate :

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

0%
$1$
0%
$\dfrac {1}{2}$
0%
$0$
0%
$none\ of\ these$

Evaluate :

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

0%
$\dfrac {3}{5}$
0%
$\dfrac {4}{5}$
0%
$\dfrac {7}{8}$
0%
$none\ of\ these$

The value of

$\displaystyle sin^{-1} \left \{ (sin\,\pi/3) \dfrac{x}{\sqrt{x^2 + k^2 - kx}} \right \} - cos^{-1} \left \{ cos\pi /6 \,\dfrac{x}{\sqrt{x^2 + k^2 - kx}} \right \}$

$\left (where \,\dfrac{k}{2} < x < 2k , \,k > 0 \right )$ is

0%
$tan^{-1} \left ( \dfrac{2x^2 + xk - k^2}{x^2 - 2xk + k^2} \right )$
0%
$tan^{-1} \left ( \dfrac{x^2 + 2xk - k^2}{x^2 - 2xk + k^2} \right )$
0%
$tan^{-1} \left ( \dfrac{x^2 + 2xk - 2k^2}{2x^2 - 2xk + 2k^2} \right )$
0%
none of these

$sin^{-1}\dfrac{5}{x}+sin^{-1}\dfrac{12}{x}=\dfrac{\pi}{2}$ , then x is equal to

0%
$\dfrac{7}{13}$
0%
$\dfrac{4}{3}$
0%
$13$
0%
$\dfrac{13}{7}$

Explanation

Put

$sin^{-1}\dfrac{5}{x}=A\implies \dfrac{5}{x}=\sin A$

$\sin^{-1}\dfrac{12}{x}=B\implies \dfrac{12}{x}=\sin B \implies A+B=\dfrac{\pi}{2}$

$\implies\sin A=\sin(\dfrac{\pi}{2}-B)=\cos B=\sqrt{1-sin^{2}B}$

$\implies\dfrac{5}{x}=\sqrt{1-\dfrac{144}{x^2}}\implies \dfrac{169}{x^2}=1$

$\implies x^2=169 \implies x=13$ [

$x = -13$ does not satisfy the given equation]

$\tan^{1}x + \tan^{1}y + \tan^{1}z = \dfrac {\pi}{2}$ , then

0%
$xy+yz+zx-xyz=0$
0%
$xy+yz+zx+xyz=0$
0%
$xy+yz+zx+1=0$
0%
$xy+yz+zx-1=0$

Explanation

$Given\,\,that \tan^{1}x + \tan^{1}y + \tan^{1}z = \dfrac {\pi}{2}$

$\implies \tan^{-1}\Bigg[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\Bigg]=\dfrac{\pi}{2}$

For this denominator of

$\Bigg[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\Bigg]$ must be zero.

hence

$xy+yz+zx-1=0$

$cos^{-1}(cos(\dfrac{5\pi}{4}))$ is given by

0%
$\dfrac{5\pi}{4}$
0%
$\dfrac{3\pi}{4}$
0%
$\dfrac{-\pi}{4}$
0%
none of these

Explanation

$cos^{-1}\bigg(cos\dfrac{5\pi}{4}\bigg)=cos^{-1}\Bigg(cos\bigg(2\pi-\dfrac{5\pi}{4}\bigg)\Bigg)=cos^{-1}\Bigg(cos\bigg(\dfrac{3\pi}{4}\bigg)\Bigg)=\dfrac{3\pi}{4}$

The value of $sin^{-1}\Bigg(cot\bigg(sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+\sec^{-1}\sqrt{2}\bigg)\Bigg)$ is

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

Explanation

We have

$sin^{-1}\Bigg(cot\bigg(sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+\sec^{-1}\sqrt{2}\bigg)\Bigg)$

$\sin^{-1}\Bigg(\cot\bigg(sin^{-1}\bigg(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\bigg)+cos^{-1}\dfrac{\sqrt{3}}{2}+\cos^{-1}\dfrac{1}{\sqrt{2}}\bigg)\Bigg)$

$\sin^{-1}[\cot(15^{\circ}+30^{\circ}+45^{\circ})]$

$\sin^{-1}[\cot(90^{\circ})]$

$\sin^{-1}(0)=0$

$\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{1}{2}\tan^{-1}x$ , then x is equal to

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

Explanation

We have

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

$\implies\,\tan^{-1}\Bigg[\dfrac{1-\tan\theta}{1+\tan\theta}\Bigg]=\dfrac{1}{2}\theta$

$\implies\,\tan^{-1}\Bigg[\dfrac{\tan\dfrac{\pi}{4}-\tan\theta}{1+\tan\dfrac{\pi}{4}\tan\theta}\Bigg]=\dfrac{1}{2}\theta$

$\implies\,\tan^{-1}\tan\bigg(\dfrac{\pi}{4}-\theta\bigg)=\dfrac{\theta}{2}$

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

$\implies\theta=\dfrac{\pi}{6}=\tan^{-1}x$

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

$tan^{-1}x+2 cot^{-1}x=\dfrac{2\pi}{3}$ , then x is equal to

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

Explanation

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

$\implies tan^{-1}x=2\bigg(\dfrac{\pi}{3}-\cot^{-1}x\bigg)=2\bigg(\dfrac{\pi}{3}-(\dfrac{\pi}{2}-\tan^{-1}x)\bigg)=2\bigg(-\dfrac{\pi}{6}+\tan^{-1}x\bigg)$

$\implies\tan^{-1}x=\dfrac{\pi}{3} \implies x=\tan\dfrac{\pi}{3}=\sqrt{3}$

$3\sin^{-1}\bigg(\dfrac{2x}{1+x^2}\bigg)-4\cos^{-1}\bigg(\dfrac{1-x^2}{1+x^2}\bigg)+2tan^{-1}\bigg(\dfrac{2x}{1-x^2}\bigg)=\dfrac{\pi}{3}$ , where

$|x|<1$ . then x is equal to

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

Explanation

Put

$x=\tan\theta$

$\therefore\,3\sin^{-1}\bigg(\dfrac{2\tan\theta}{1+\tan^2\theta}\bigg)-4\cos^{-1}\bigg(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\bigg)+2\tan^{-1}\bigg(\dfrac{2\,\tan\theta}{1-\tan^2\theta}\bigg)=\dfrac{\pi}{3}$

$\implies\,3\sin^{-1}(\sin2\theta)-4\cos^{-1}(\cos2\theta)+2\tan^{-1}(\tan2\theta)=\dfrac{\pi}{3}$

$\implies\,3(2\theta)-4(2\theta)+2(2\theta)=\dfrac{\pi}{3}\,\implies2\theta=\dfrac{\pi}{3}\implies\theta=\dfrac{\pi}{6}\implies\,x=\tan \dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}$

The principal value of

$\sin^{-1}(\sin 10)$ is

0%
$10$
0%
$10-3\pi$
0%
$3\pi-10$
0%
none of these

Explanation

$\sin^{-1}(\sin 10)=\sin^{-1}[\sin (3\pi-10)]=3\pi-10$

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

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

Explanation

The given equation can be written as

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

$\implies 3(15^{\circ})=\tan^{-1}\dfrac{\dfrac{1}{x}+\dfrac{1}{3}}{1-\dfrac{1}{x}\dfrac{1}{3}}\,\implies1=\dfrac{3+x}{3x-1}\implies x=2$

The value

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ is equal to

0%
$\cos^{-1}\Bigg[\dfrac{a\cos\theta+b}{b\cos\theta+a}\Bigg]$
0%
$\cos^{-1}\Bigg[\dfrac{b\cos\theta+a}{a\cos\theta+b}\Bigg]$
0%
$\cos^{-1}\Bigg[\dfrac{a\cos\theta}{b\cos\theta+a}\Bigg]$
0%
$\cos^{-1}\Bigg[\dfrac{b\cos\theta}{a\cos\theta+b}\Bigg]$

Explanation

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

$cos^{-1}\Bigg[\dfrac{1-\bigg(\dfrac{a-b}{a+b}\bigg)\tan^2\dfrac{\theta}{2}}{1+\bigg(\dfrac{a-b}{a+b}\bigg)\tan^2\dfrac{\theta}{2}}\Bigg]$

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

$cos^{-1}\Bigg[\dfrac{(a+b)-(a-b)\tan^2\dfrac{\theta}{2}}{(a+b)+{a-b}\tan^2\dfrac{\theta}{2}}\Bigg]$

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

$cos^{-1}\Bigg[\dfrac{a(1-\tan^2\dfrac{\theta}{2})+b(1+\tan^2\dfrac{\theta}{2})}{a(1+\tan^2\dfrac{\theta}{2})+b(1-\tan^2\dfrac{\theta}{2})}\Bigg]$

$2tan^{-1}\Bigg[\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac{\theta}{2}\Bigg]$ =

$cos^{-1}\Bigg[\dfrac{\dfrac{a(1-\tan^2\dfrac{\theta}{2})}{(1+\tan^2\dfrac{\theta}{2})}+b}{a+b\dfrac{(1-\tan^2\dfrac{\theta}{2})}{(1+\tan^2\dfrac{\theta}{2})}}\Bigg]$

$=\cos^{-1}\Bigg[\dfrac{a\cos\theta+b}{b\cos\theta+a}\Bigg]$

$x=\sec^{-1}\Bigg(x+\dfrac{1}{x}\Bigg)+\sec^{-1}\Bigg(y+\dfrac{1}{y}\Bigg)$ where xy < 0, then the possible values of z is (are)

0%
$\dfrac{8\pi}{10}$
0%
$\dfrac{7\pi}{10}$
0%
$\dfrac{9\pi}{10}$
0%
$\dfrac{21\pi}{20}$

Explanation

$xy<0\Rightarrow x+\dfrac{1}{x}\geq2,y+\dfrac{1}{y}\leq-2$
or

$x+\dfrac{1}{x}\leq-2,y+\dfrac{1}{y}\geq2$

$x+\dfrac{1}{x}\geq2\Rightarrow\sec^{-1}\Bigg(x+\dfrac{1}{x}\Bigg)\in[\dfrac{\pi}{3},\dfrac{\pi}{3})$

$y+\dfrac{1}{y}\leq-2\Rightarrow\sec^{-1}\Bigg(y+\dfrac{1}{y}\Bigg)\in(\dfrac{\pi}{2},\dfrac{2\pi}{3}] \,\Rightarrow z\in\bigg(\dfrac{5\pi}{6},\dfrac{7\pi}{6}\bigg)$

$tan^{-1}(\sin^2\theta-2sin\theta+3)+\cot^{-1}(5^{\sec^2y}+1)=\dfrac{\pi}{2}$ , then the value of

$cos^2\theta-\sin\theta$ is equal to

0%
$0$
0%
$-1$
0%
$1$
0%
none of these

Explanation

From the given equation

$\sin^2\theta-2sin\theta+3=5^{\sec^2y}+1$ , we get

$(\sin\theta-1)^2+2=5^{\sec^2y}+1$

$L.H.S.\leq6 , R.H.S.\geq6$

Possible solution is

$sin\theta=-1$ when L.H.S. = R.H.S.

$\Rightarrow cos^2\theta=0\,\Rightarrow cos^2\theta-\sin\theta=1$

If f(x)=

$sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}x-\dfrac{1}{2}\sqrt{1-x^2}\Bigg),-\dfrac{1}{2}\leq x\leq1$ , then f(x) is equal to

0%
$\sin^{-1}\bigg(\dfrac{1}{2}\bigg)-sin^{-1}x$
0%
$\sin^{-1}x-\dfrac{\pi}{6}$
0%
$\sin^{-1}x+\dfrac{\pi}{6}$
0%
none of these

Explanation

Let

$x=\sin\theta$ where

$\dfrac{-1}{2}\leq x\leq1\implies\,\dfrac{-\pi}{6}\leq\theta\leq\dfrac{\pi}{2}$

Then

$f(x)=sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}x-\dfrac{1}{2}\sqrt{1-x^2}\Bigg)$

$=sin^{-1}\Bigg(\dfrac{\sqrt{3}}{2}\sin\theta-\dfrac{1}{2}\cos\theta\Bigg)$

$=sin^{-1}\Bigg(\sin\bigg(\theta-\dfrac{\pi}{6}\bigg)\Bigg)$

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

$cot^{-1}(\sqrt{\cos a})- tan^{-1}(\sqrt{\cos a})=x$ , then

$\sin x$ is

0%
$tan^2\dfrac{\alpha}{2}$
0%
$cot^2\dfrac{\alpha}{2}$
0%
$\tan \alpha$
0%
$cot\dfrac{\alpha}{2}$

Explanation

$cot^{-1}(\sqrt{\cos a})- tan^{-1}(\sqrt{\cos a})=x$

$tan^{-1}(\dfrac{1}{\sqrt{\cos a}})- tan^{-1}(\sqrt{\cos a})=x$

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

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

$\dfrac{1-\cos a}{2\sqrt{\cos a}}=\tan x$

$\dfrac{2\sqrt{\cos a}}{1-\cos a}=\cot x$

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

$tan^2\dfrac{\alpha}{2}$

$2tan^{-1}(-2)$ is equal to

0%
$-\cos^{-1}\bigg(\dfrac{-3}{5}\bigg)$
0%
$-\pi+\cos^{-1}\bigg(\dfrac{3}{5}\bigg)$
0%
$-\dfrac{\pi}{2}+\tan^{-1}\bigg(\dfrac{-3}{4}\bigg)$
0%
$-\pi+\cot^{-1}\bigg(\dfrac{-3}{4}\bigg)$

Explanation

Let

$tan^{-1}(-2)=\theta\,\Rightarrow\,\tan\theta=-2\,\Rightarrow\,\theta=(\dfrac{-\pi}{2},0)$

$\Rightarrow2\theta=(-\pi,0)$

$\cos(-2\theta)=\cos(2\theta)=\dfrac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\dfrac{-3}{5}$

$\Rightarrow -2\theta=\cos^{-1}\dfrac{-3}{5}=\pi-\cos^{-1}\dfrac{3}{5}$

$\Rightarrow 2\theta=-\pi+\cos^{-1}\dfrac{3}{5}=-\pi+\tan^{-1}\dfrac{4}{3}=-\pi+\cot^{-1}\dfrac{3}{4}=-\dfrac{\pi}{2}-tan^{-1}\dfrac{3}{4}$

$=-\dfrac{\pi}{2}+tan^{-1}\dfrac{-3}{4}$

The value of

$\underset{|x|\rightarrow\infty}{lim}cos(tan^{-1}(sin(tan^{-1}x)))$ is equal to

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

Explanation

$\underset{|x|\rightarrow\infty}{lim}cos(tan^{-1}(sin(tan^{-1}x)))=cos(tan^{-1}(sin(tan^{-1}\infty)))$

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

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

$=cos(\dfrac{\pi}{4})=\dfrac{1}{\sqrt{2}}$

$x\in\bigg(\dfrac{-\pi}{2},\dfrac{\pi}{2}\bigg)$ , then the value

$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\sin 2x}{5+3\cos 2x}\Bigg)$ is

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

Explanation

$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\sin 2x}{5+3\cos 2x}\Bigg)$ =

$\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)$ +

$\tan^{-1}\Bigg(\dfrac{\dfrac{6\tan x}{1+tan^{2}x}}{5+3\dfrac{1-tan^{2}x}{1+tan^{2}x}}\Bigg)$

$=\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{6\tan x}{8+2\tan^2x}\Bigg)$

$=\tan^{-1}\Bigg(\dfrac{\tan x}{4}\Bigg)+\tan^{-1}\Bigg(\dfrac{3\tan x}{4+\tan^2x}\Bigg)$

$=\tan^{-1}\Bigg(\dfrac{\dfrac{\tan x}{4}+\dfrac{3\tan x}{4+\tan^2x}}{1-\dfrac{3\tan^2x}{4(4+\tan^2x)}}\Bigg)$

$=\tan^{-1}\Bigg(\dfrac{16\tan x+tan^{3}x}{16+tan^2x}\Bigg)$

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

$\alpha,\beta \,\,and\,\, \gamma$ are the roots of

$\displaystyle \tan^{-1} \left ( x-1 \right ) +\tan^{-1} x + \tan^{-1} \left ( x+1 \right ) = \tan^{-1} 3x$ , then

0%
$\alpha+\beta+ \gamma=0$
0%
$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{-1}{4}$
0%
$\alpha\beta\gamma=1$
0%
$|\alpha-\beta|_{max}=1$

Explanation

Given,

$\displaystyle \tan^{-1} \left ( x-1 \right ) +\tan^{-1} x + \tan^{-1} \left ( x+1 \right ) = \tan^{-1} 3x$

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

$\Rightarrow\tan^{-1}\Bigg[\dfrac{(x-1)+x}{1-(x-1)x}\Bigg]=\tan^{-1}\Bigg[\dfrac{3x-(x+1)}{1+(x+1)3x}\Bigg]$

$\Rightarrow\Bigg[\dfrac{2x-1}{1-x^2+x}\Bigg]=\Bigg[\dfrac{2x-1}{1+3x^2+3x}\Bigg]$

$\Rightarrow(1 - x^2 + x) (2x - 1) = ( 1 + 3x^2 + 3x) (2x- 1)$

$\Rightarrow x=0,\pm \dfrac{1}{2}$

Given that

$\alpha,\beta \,\,and\,\, \gamma$ are the roots of given equation.

$\alpha+\beta+ \gamma=\dfrac 12-\dfrac12+0=0$

$\alpha\beta+\beta\gamma+\gamma\alpha=0+\left(\dfrac{1}{2}\times \dfrac{-1}{2}\right )+0=\dfrac{-1}{4}$

$\alpha\beta\gamma=0$

$|\alpha-\beta|_{max}=\left|\dfrac{-1}2-\dfrac 12\right|=1$

If f(x) =

$(\sin^{-1}x)^2 +(\cos^{-1}x)^2$ , then.

0%
f(x) has the least value of $\dfrac{\pi^2}{8}$
0%
f(x) has the least value of $\dfrac{5\pi^2}{8}$
0%
f(x) has the least value of $\dfrac{\pi^2}{16}$
0%
f(x) has the least value of $\dfrac{5\pi^2}{4}$

Explanation

Let

$f (x) = (\sin^{-1}x)^2 +(\cos^{-1}x)^2$ ,

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

$=\dfrac{\pi^2}{4}- 2\sin^{-1}x\Bigg[\dfrac{\pi}{2}-\sin^{-1}x\Bigg]$

$=\dfrac{\pi^2}{4}- \pi\sin^{-1}x2(\sin^{-1}x)^2$

$=2\Bigg(\sin^{-1}x-\dfrac{\pi}{4}\Bigg)^2+2\Bigg[\dfrac{\pi^2}{16}\Bigg]$

Now,

$-\dfrac{\pi}{2}\leq sin^{-1}x\leq\dfrac{\pi}{2}$

$\Rightarrow-\dfrac{3\pi}{4}\leq sin^{-1}x-\dfrac{\pi}{4}\leq\dfrac{\pi}{4}$

$\Rightarrow0\leq \bigg(sin^{-1}x-\dfrac{\pi}{4}\bigg)^2\leq\dfrac{9\pi^2}{16}$

$\Rightarrow0\leq 2\bigg(sin^{-1}x-\dfrac{\pi}{4}\bigg)^2\leq\dfrac{9\pi^2}{8}$

$\Rightarrow\dfrac{\pi^2}{8}\leq 2\bigg(sin^{-1}x-\dfrac{\pi}{4}\bigg)^2+\dfrac{\pi^2}{8}\leq\dfrac{5\pi^2}{4}$

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

$\sin 2x=\cos 2y$ , then

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$x=\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$
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$y=-\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$
0%
$x=\dfrac{\pi}{12}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$
0%
$y=-\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$

Explanation

For the given equation

$0\leq x,y\leq1$

Also,

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

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

$\Rightarrow x=\sqrt{1-y^2} \,\Rightarrow\, x^2+y^2=1$ (i)

Again,

$\sin 2x = \cos 2y$

$\Rightarrow\cos\bigg(\dfrac{\pi}{2}-2x\bigg)=\cos 2y$

$\dfrac{\pi}{2}-2x=2ny\pm2y,where\,n \in I$

$x\pm y=\dfrac{\pi}{4}-n\pi$ (ii)

From Eqs. (i) and (ii), we get

$x=\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$ and

$y=-\dfrac{\pi}{8}+\sqrt{\dfrac{1}{2}-\dfrac{\pi^2}{64}}$

$cos^{-1}x+cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{1}{2}\sqrt{3-3x^2}\Bigg]$ is equal to

0%
$\dfrac{\pi}{3}$ for $x\in\Bigg[\dfrac{1}{2},1\Bigg]$
0%
$\dfrac{\pi}{3}$ for $x\in\Bigg[0,\dfrac{1}{2}\Bigg]$
0%
$2cos^{-1}x-cos^{-1}\dfrac{1}{2}$ for $x\in\Bigg[\dfrac{1}{2},1\Bigg]$
0%
$2cos^{-1}x-cos^{-1}\dfrac{1}{2}$ for $x\in\Bigg[0,\dfrac{1}{2}\Bigg]$

Explanation

Case 1:

$0\leq x\leq\dfrac{1}{2}$ , then

$cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{1}{2}\sqrt{3-3x^2}\Bigg]=cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{\sqrt{3}}{2}\sqrt{1-x^2}\Bigg]=cos^{-1}x-cos^{-1}\dfrac{1}{2}$

Case 2:

$\dfrac{1}{2}\leq x\leq1$ , then

$cos^{-1}\Bigg[\dfrac{x}{2}+\dfrac{1}{2}\sqrt{3-3x^2}\Bigg]=-cos^{-1}x+cos^{-1}\dfrac{1}{2}$

The domain of the function

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

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$[0,1]$
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$[-1,1]$
0%
$(-1,1)$
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$[0,\pi]$

Explanation

we have,

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

$\because -1 \le 2x -1 \le1$

$\Rightarrow 0 \le 2x \le 2$

$\Rightarrow 0 \le x \le 1$

$\therefore x \in [0,1]$

$\cos \left(\sin ^{-1}\dfrac{2}{5} + \cos ^{-1} x\right)=0$ then

$x$ is equal to

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$\dfrac{1}{5}$
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$\dfrac{2}{5}$
0%
$0$
0%
$1$

Explanation

we have

$\cos \left(\sin ^{-1}\dfrac{2}{5} + \cos ^{-1} x\right)=0$

$\Rightarrow \sin ^{-1}\dfrac{2}{5}+ \cos ^{-1}x= \cos^{-1}0$

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

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

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

$\Rightarrow \cos ^{-1}x=\cos ^{-1} \dfrac{2}{5} \ \ [\because \cos^{-1}x+ \sin ^{-1}x=\dfrac{\pi}{2}]$

$\therefore x=\dfrac{2}{5}$

The domain of trigonometric function can be restricted to any one of their branch (not necessarily principle value) in order to obtain their inverse functions.

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True
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False

The minimum value of

$n$ for which

$\tan^{-1}\dfrac {n}{\pi} > \dfrac {\pi}{4}, n\ \in \ N$ is valid is

$5$ .

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True
0%
False

Explanation

False

$\because \tan^{-1} \dfrac {n}{\pi} > \dfrac {\pi}{4} \Rightarrow \dfrac {n}{\pi} > \tan \dfrac {\pi}{4}$

$\Rightarrow \dfrac {n}{\pi} > 1 \left[\because \tan \dfrac {\pi}{4} =1\right]$

$\Rightarrow n > \pi$
So, the minimum value of

$n$ is

$4 [\because n\in N\ and\ \pi=3.14....]$

All trigonometric functions have inverse over their respective domains.

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True
0%
False

The value of

$\cot \left[cos^{-1}\left(\dfrac{7}{25}\right)\right]$ is

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$\dfrac{25}{24}$
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$\dfrac{25}{7}$
0%
$\dfrac {24} {25}$
0%
$\dfrac {7} {24}$

Explanation

we have

$\cot \left[cos^{-1}\left(\dfrac{7}{25}\right)\right]$
Let

$\cos ^{-1}\dfrac{7}{25}=x$

$\Rightarrow \cos x=\dfrac{7}{25}$
$$\therefore \sin x=\sqrt{1-\cos^2x=}\sqrt{1-\left(\dfrac{7}
{25}\right)^2}$$

$=\sqrt{\dfrac{625-49}{625}}=\dfrac{24}{25}$
$$\therefore \cot x=\dfrac{\cos x}{\sin x}= \dfrac{\dfrac{7}
{25}}{\dfrac{24}{25}}=\dfrac{7}{24}$$

$\Rightarrow x=\cot^{-1} \left(\dfrac{7}{24} \right)=\cot^{-1} \left(\dfrac{7}{25}\right)$

$$\therefore \cot \left(\cos^{-1}\dfrac{7}{25}\right)= \cot
\left(\cos^{-1}\dfrac{7}{24}\right)=\dfrac{7}{24} \ \ \left
[\because \cos^{-1}\dfrac{7}{24}=\cos^{-1}\dfrac{7}
{25}\right]$$
1656728_1795509_ans_4e362b3b1af34e6481c5b812a1d164f7.png

1656728_1795509_ans_4e362b3b1af34e6481c5b812a1d164f7.png

$|x| \le 1$ , then

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

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

Explanation

We have,

$|x| \le 1$ , then

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

Let

$x=\tan \theta$

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

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

$2 \theta +\sin^{-1} \sin 2 \theta \ \left[\because \sin 2 \theta =\dfrac{2 \tan \theta}{1+ \tan^2 \theta}\right]$

$2 \theta +2 \theta \left[\because \sin^{-1} (\sin x)=x \right]$

$4 \theta \left[\because \theta \tan^{-1} \right]4 \tan^{-1} x$

The value of the expression

$2 \sec ^{-1}2+ \sin^{-1}\left(\dfrac{1}{2}\right)$ is

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{5\pi}{6}$
0%
$\dfrac{7\pi}{6}$
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$1$

Explanation

We have

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

$2 \sec^{-1}(\sec \dfrac{\pi}{3})+\sin^{-1}(\sin \dfrac{\pi}{6})$

$=2.\dfrac{\pi}{3}+\dfrac{\pi}{6} \ \ [\because \sec^{-1} (\sec x)=x$ and

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

$=\dfrac{4 \pi+\pi}{6}=\dfrac{5\pi}{6}$

The value of

$\sin [2 \tan ^{-1} (.75)]$ is equal to

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$0.75$
0%
$1.5$
0%
$0.96$
0%
$\sin 1.5$

Explanation

we have,

$\sin [2 \tan ^{-1} (.75)] = \sin \left(2 \tan ^{-1}\dfrac{3}{4}\right) [0.75=\dfrac{75}{100}=\dfrac{3}{4}$

$=\sin \left(\sin ^{-1}\dfrac{2.\dfrac{3}{4}}{1+\dfrac{9}{16}}\right)=\sin \left[\sin ^{-1} \dfrac{3/2}{25/16}\right]$

$=\sin \left[\sin^{-1} \left(\dfrac{48}{50}\right)\right]=\sin \left[\sin^{-1} \left(\dfrac{48}{50}\right)\right]=\dfrac{24}{25}=0.96$

The value of

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

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

Explanation

We have,

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

$=\cos ^{-1}\cos \left( 2 \pi \dfrac{ \pi}{2}\right) \ \ \ \left[\because \cos \left(2 \pi -\dfrac{\pi}{2}\right)\right]= \cos \dfrac{\pi}{2}$

$=\cos ^{-1}\cos \left( \dfrac{ \pi}{2}\right) =\dfrac{ \pi}{2} \ \ \ \left\{\because \cos^{-1} (\cos x)=x, x \in [0,\pi] \right\}$
Note remember that,

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

$\because \dfrac{3 \pi }{2} \in (0, \pi)$

The value of the expression

$(\cos^{-1}x)^2$ is equal to

$\sec^2 x$ .

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True
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False

Explanation

False

$\because [\cos^{-1}x]^2 =\left[\sec^{-1}\dfrac 1x \right]^2\neq \sec^2 x$

$\cos ^{-1} \alpha +\cos ^{-1} \beta +\cos ^{-1} \gamma = 3 \pi$ , then

$\alpha (\beta + \gamma) + \beta (\gamma + \alpha) + \gamma (\alpha + \beta)$

0%
$0$
0%
$1$
0%
$6$
0%
$12$

Explanation

we have,

$\cos ^{-1} \alpha +\cos ^{-1} \beta +\cos ^{-1} \gamma = 3 \pi$

$\alpha (\beta + \gamma) + \beta (\gamma + \alpha) + \gamma (\alpha + \beta)$

we know that ,

$0 \le \cos^{-1} x \le \pi$
$$\Rightarrow \cos^{-1} \alpha +\cos^{-1} \beta +\cos^{-1}
\gamma = 3 \pi$$
If and only if, $$\cos^{-1}\alpha=\cos^{-1} \beta =\cos^{-1}
\gamma =\pi$$

$\Rightarrow \cos \pi=\alpha=\beta=\gamma$

$\Rightarrow -1=\alpha=\beta=\gamma$
$$\therefore \alpha (\beta + \gamma) + \beta (\gamma +
\alpha)+ \gamma (\alpha + \beta)$$

$=-1(-1-1)-1(1-1)-1(-1-1)$

$=2+2+2=6$

The domain of the function defined by

$f(x)=\sin^{-1}x+\cos x$ is

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$[-1, 1]$
0%
$[-1, \pi +1]$
0%
$( -\infty, \infty)$
0%
$\phi$

Explanation

(A) is the correct answer.

The domain of

$\cos x$ is

$R$ and the domain of

$\sin^{-1}$ is

$[-1, 1]$ .

$\therefore$ The domain of

$f(x)=\cos x+\sin^{-1}x$ is

$R \cap [-1, 1]$ i.e.,

$[-1, 1]$

$\sin^{-1}x+\sin^{-1}y=\dfrac{\pi}{2}$ , then the value of

$\cos^{-1}x+\cos^{-1}y$ is

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

Explanation

(A) is the correct answer.

Given

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

We know that

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

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

Hence,

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

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 6 - 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 6 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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