If $$\theta =\cot ^{ -1 }{ \sqrt { \cos { x } } -\tan ^{ -1 }{ \sqrt { \cos { x } } } } $$, then $$\sin { \theta } =$$

$$\tan { \frac { 1 }{ 2 } } x$$

$$\tan ^{ 2 }{ \left( \frac { x }{ 2 } \right) } $$

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

If $$\cos ^ { - 1 } ( x / a ) + \cos ^ { - 1 } ( y / b ) = \alpha ,$$ Then $$x ^ { 2 } / a ^ { 2 } + y ^ { 2 } / b ^ { 2 }$$ is equal to:

$$( 2 x y / a b ) \cos \alpha + \sin ^ { 2 } \alpha$$

$$( 2 x y / a b ) \sin \alpha + \cos ^ { 2 } \alpha$$

$$( 2 x y / a b ) \cos ^ { 2 } \alpha + \sin \alpha$$

$$( 2 x y / a b ) \sin ^ { 2 } \alpha + \cos \alpha$$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 10

If for

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

$x=$

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

The value of a for which

$ax+sec^{ -1 }\sqrt { { 2x }^{ 2 }-{ x }^{ 4 } } +cose^{ -1 }\sqrt { { 2x }^{ 2 }-{ x }^{ 4 } } =0$

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

The sum of the solution of the equation

$2\sin^{-1}\sqrt {x^{2}+x+1}+\cos^{-1}\sqrt {x^{2}+x}\dfrac {3\pi}{2}$ is

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

$\tan^{-1}{\left(\dfrac{3a^{2}x^{3}}{a^{3}-3ax^{2}}\right)}=k\tan^{-1}{\left(\dfrac{x}{a}\right)}$ , then

$k=$

0%
$2$
0%
$3$
0%
$-2$
0%
$4$

${ cos }^{ -1 }\sqrt { p } +{ cos }^{ -1 }\sqrt { 1-p } +{ cos }^{ -1 }\sqrt { 1-q } =\frac { 3\pi }{ 4 }$ then the value of q is equal to:

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

The value of

$\sin ^{-1}[\cot{(\sin ^{-1}\sqrt{(\frac{2 - \sqrt 3}{4})} + \cos ^{-1}\frac{\sqrt {12}}{4})} + \sec ^{-1}\sqrt 2]$

0%
0
0%
$\frac{\pi}{4}$
0%
$\frac{\pi}{6}$
0%
$\frac{\pi}{2}$

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

${ cot }^{ -1 }\left( { 4 }^{ { sec }^{ 2 } }+1 \right) =\frac { \pi }{ 2 }$ has solution for some

$\theta$ and

$\phi$ then

0%
$sin\theta =-1$
0%
$cos\phi =-1$
0%
$cos\phi =1$
0%
$sin\theta =1$

The sum of roots of the equation

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

0%
$\dfrac{7}{3}$
0%
$3$
0%
$4$
0%
$5$

$\sin^{-1} x+\sin^{-1} y=\dfrac{\pi}{3}$ then the value of

$\cos^{-1}x+\cos^{-1}y$ is equal to which of the following :

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

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

0%
$\frac{\pi}{2}$
0%
$\frac{\pi}{3}$
0%
$\frac{\pi}{2}$
0%
$\frac{\pi}{4}$ or $\frac{3 \pi}{4}$

${\sin ^{ - 1}}x + {\cos ^{ - 1}}y = \frac{{2\pi }}{5},$ then

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

0%
$\frac{{2\pi }}{5}$
0%
$\frac{{3\pi }}{5}$
0%
$\frac{{4\pi }}{5}$
0%
$\frac{{3\pi }}{10}$

The value of x satisfying

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

0%
0
0%
1/2
0%
1
0%
2

Solve

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

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

$x = {\sin ^{ - 1}}\left( {\sin 10} \right)\,\,and\,\,y = {\cos ^{ - 1}}\left( {\cos 10} \right),\,\,then\,\,y - x$ is equal to:

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

$x = {\sin ^{ - 1}}\left( {\sin 10} \right)\,\,and\,\,y = {\cos ^{ - 1}}\left( {\cos 10} \right)$

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

$\theta \epsilon [4\pi ,5\pi ]$ , then

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

0%
$-4\pi +\theta$
0%
$5\pi -\theta$
0%
$4\pi -\theta$
0%
$\theta -5\pi$

The solution set of the equation

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

0%
[-1,1] - {0}
0%
(0,1] U {-1]
0%
[-1,0) U {1}
0%
[-1,1]

$[\sin^{-1}(\cos ^{-1}(\sin ^{-1}(\tan ^{-1}x)))]=1$ , where

$\left[ \bullet \right]$ denotes the greatest integer function, then

$x\in$

0%
$[\tan \sin \cos 1,\tan \sin \cos \sin 1]$
0%
$(\tan \sin \cos 1,\tan \sin \cos \sin 1)$
0%
[-1, 1]
0%
None of these

Number of solutions of the equation

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

0%
$one$
0%
$Two$
0%
$Three$
0%
$Zero$

Solution of

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

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

${\cos ^{ - 1}}\left( {\cos \;x} \right) = [x],[.]\;denotes\;the\;greatest\;\operatorname{int} eger\;function,\;is\;$

0%
$2\pi + 3$
0%
$\pi + 3$
0%
$\pi - 3$
0%
$2\pi - 3$

$\alpha$ is the only real root of the equation

$\displaystyle x^3 + bx^2 + c = 0$ (

$b$ <

$c$ ) then the value of

$\displaystyle Tan^{1}\alpha + Tan^{1}\left ( \dfrac{1}{\alpha} \right )$ =

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

The number of solution for the equation

$\cos^{-1}(1-x)+m\ \cos^{-1}x=\dfrac {n\pi}{2}$ where

$m > 0,n \le 0$ is

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

The value of

${ sin }^{ -1 }\left[ cot\left( { sin }^{ -1 }\sqrt { \left( \cfrac { 2-\sqrt { 3 } }{ 4 } \right) } +{ cos }^{ -1 }\left( \cfrac { \sqrt { 12 } }{ 4 } \right) +{ sec }^{ -1 }\sqrt { 2 } \right) \right]$ is :

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

$x$ =

$Tan^{1}(1)$ +

$Cos^{1}(\dfrac{1}{2})$ +

$Sin^{1}(\dfrac{1}{2})$ and

$y$ =

$cos\left [ \dfrac{1}{2}Cos^{1}(1/8) \right ]$ then

0%
$x$ = $2\, \pi\, y$
0%
$y$ = $3\, \pi\, x$
0%
$x$ = $\pi\, y$
0%
$y$ = $\pi\, x$

Solve the equation :

$4 \tan ^ { - 1 } \dfrac { 1 } { 5 } - \tan ^ { - 1 } \dfrac { 1 } { 239 } =?$

0%
$\pi$
0%
$\pi / 2$
0%
$\pi / 3$
0%
$\pi / 4$

Let

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

$\frac { \pi } { 2 }$ is equal to:

0%
f(-2)
0%
$f \left( k ^ { 2 } - 2 k + 3 \right) , k \in R$
0%
$\mathrm { f } \left( \frac { 1 } { 1 + \mathrm { k } ^ { 2 } } \right) , \mathrm { k } \in \mathrm { R }$
0%
none

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

${ cos }^{ -1 }x+{ cos }^{ -1 }y=$

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

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

$x$ =

0%
$-1$
0%
$\dfrac{1}{6}$
0%
$-1, \dfrac{1}{6}$
0%
None of these

$A=\tan {-1}(\frac {x\sqrt 3}{2k-x})$ and

$B=\tan ^{-1}(\frac {2x-k}{k\sqrt 3})$ , then the value of A-B is

0%
$0^o$
0%
$45^o$
0%
$60^o$
0%
$30^o$

$\displaystyle Sin^{1}(x - \dfrac{x^2}{2} + \dfrac{x^3}{4} - .......) + Cos^{1}(x^2 - \dfrac{x^4}{2} + \dfrac{x^6}{4} - ......) = \dfrac{\pi}{2}$ for 0 <

$\left | x \right |$ <

$\sqrt{2}$ then

$x$ =

0%
1/2
0%
1
0%
1/2
0%
1

$\sin \cot ^{-1} \tan \cos ^{-1} x$ is always equal to

0%
x
0%
$\sqrt{1-x^{2}}$
0%
$\frac{1}{x}$
0%
None of these

$3tan^{ -1 }x\quad =\quad tan^{ -1 }\{ \dfrac { 3x-x^{ 3 } }{ 1-3x^{ 2 } } \} ,\quad then\quad x\quad belong\quad to\quad$

0%
[-1,1]
0%
(-1,1)
0%
$\{ -\dfrac { 1 }{ \sqrt { 3 } } ,\dfrac { 1 }{ \sqrt { 3 } } \}$
0%
None of these

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

$B=\tan {-1}(\frac {2x-k}{k\sqrt 3})$ , then the value of

$A-B$ is

0%
$0^o$
0%
$45^o$
0%
$60^o$
0%
$30^o$

Value of

$\sin ^{ -1 }{ \frac { 3 }{ \sqrt { 13 } } +\cos ^{ -1 }{ \frac { 11 }{ \sqrt { 146 } } } +\cot ^{ -1 }{ \sqrt { 3 } } }$ is

0%
$\pi$
0%
$\pi$ /2
0%
5 $\pi$ /12
0%
$\pi$ /3

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

$B=3{\sin}^{-1}\left(1/3\right)+{sin}^{-1}\left(3/5\right)$ , then

0%
$A=B$
0%
$A<B$
0%
$A>B$
0%
$none \of \these$

The least positive integer n for which

$\left( \frac { 1+i }{ 1-i } \right) ^{ n }=\frac { 2 }{ \pi } \left( \sec ^{ -1 }{ \frac { 1 }{ x } +\sin ^{ -1 }{ x } } \right) \left( where,\quad x\neq 0,-1\le x\ge 1\quad and\quad i=\sqrt { -1 } \right)$ , is

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

$2sin^{-1}( \frac {3}{5})-cos^{-1}(\frac{5}{13})=cos^{-1} (\lambda)$ , then

$\lambda$ is eual to

0%
323/225
0%
223/325
0%
323/325
0%
123/125

the number o real soluttion of

${ tan }^{ -1 }\sqrt { x(x+1) } +{ sin }^{ -1 }\sqrt { { x }^{ 2 }+x+1 } =\frac { \pi }{ 2 } is$

0%
zero
0%
one
0%
two
0%
infinit

The numerical value of

$cot\left( { 2sin }^{ -1 }\dfrac { 3 }{ 5 } +{ cos }^{ -1 }\dfrac { 3 }{ 5 } \right)$ is

0%
$\dfrac { -4 }{ 3 }$
0%
$\dfrac { -3 }{ 4 }$
0%
$\dfrac { 3 }{ 4 }$
0%
$\dfrac { 4 }{ 3 }$

The value of

${\sin}^{-1}\left( \sin12 \right )+{\sin}^{-1}\left( \cos12 \right )=$

0%
$0$
0%
$24-2\pi$
0%
$4\pi-24$
0%
$8\pi$

$\cot^{ -1 } 3 + \cot^{ -1 } 7 + \cot^{ -1 } 13 + n terms =$

0%
$\tan^{ -1 }\left(\dfrac { n }{ n + 2 }\right)$
0%
$\tan^{ -1 }\left(\dfrac { n + 2 }{ n }\right)$
0%
$\tan^{ -1 }\left(\dfrac { 3n }{ 4n + 1 }\right)$
0%
$\tan^{ -1 }\left(\dfrac { 4n + 1 }{ 3n }\right)$

$\theta =\cot ^{ -1 }{ \sqrt { \cos { x } } -\tan ^{ -1 }{ \sqrt { \cos { x } } } }$ , then

$\sin { \theta } =$

0%
$\tan { \frac { 1 }{ 2 } } x$
0%
$\tan ^{ 2 }{ \left( \frac { x }{ 2 } \right) }$
0%
$\frac { 1 }{ 2 } \tan ^{ -1 }{ \left( \frac { x }{ 2 } \right) }$
0%
none of these

$\sin^{ -1 }\left(\dfrac { 2a }{ 1 + a^{ 2 } }\right) + \cos^{ -1 }\left(\dfrac { 1-a^{ 2 } }{ 1 + a^{ 2 } }\right) = \tan^{ -1 } \left(\dfrac { 2x }{ 1 x^{ 2 } }\right)$ where,

$a, x \epsilon \left( 0, 1\right)$ then the value of

$x$ is

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

$\tan^{-1}{\left(\cfrac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)}=\alpha,\left(\alpha\in\left[0,\cfrac{\pi}{4}\right)\right]$ yhen

$x^2$ is equal to

0%
$\cos{2\alpha}$
0%
$\tan{2\alpha}$
0%
$\sin{2\alpha}$
0%
None of these

If If

$x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = r ^ { 2 } ,$ then

$tan^ { - 1 } \left( \frac { x y } { z r } \right) + \tan ^ { - 1 } \left( \frac { y z } { x r } \right) + \tan ^ { - 1 } \left( \frac { x z } { y r } \right)$ =

0%
$\pi$
0%
$\frac { \pi } { 2 }$
0%
0
0%
none of these.

$cos^{-1}x - cos^{-1}\dfrac{y}{2} = \alpha$ , then

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

0%
$2\, sin\, 2\alpha$
0%
$4$
0%
$4\, sin^2\alpha$
0%
$-4\, sin^2\alpha$

If a is a real of the equation

${ x }^{ 3 }+3x-tan2=0$ then

${ cot }^{ -1 }a+{ cot }^{ -1 }\dfrac { 1 }{ a } -\dfrac { x }{ 2 }$ can be equal to

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

$\cos ^ { - 1 } ( x / a ) + \cos ^ { - 1 } ( y / b ) = \alpha ,$ Then

$x ^ { 2 } / a ^ { 2 } + y ^ { 2 } / b ^ { 2 }$ is equal to:

0%
$( 2 x y / a b ) \cos \alpha + \sin ^ { 2 } \alpha$
0%
$( 2 x y / a b ) \sin \alpha + \cos ^ { 2 } \alpha$
0%
$( 2 x y / a b ) \cos ^ { 2 } \alpha + \sin \alpha$
0%
$( 2 x y / a b ) \sin ^ { 2 } \alpha + \cos \alpha$

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

0%
maximum value of ${ x }^{ 2 }+{ y }^{ 2 }$ is $\dfrac { 49 }{ 3 }$
0%
maximum value of ${ x }^{ 2 }+{ y }^{ 2 }$ is 4
0%
maximum value of ${ x }^{ 2 }+{ y }^{ 2 }$ is $\quad \dfrac { 1 }{ 2 }$
0%
maximum value of ${ x }^{ 2 }+{ y }^{ 2 }$ is 3

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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