If $$(cot^{-1}x)^{2}-3(cot^{-1}x)+2 > 0,$$ then x lies in

$$(-\infty , cot 2)\cup (cot 1,\infty )$$

$$(-\infty , cot 1)\cup (cot 2,\infty )$$

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

$$\cfrac { 1 }{ 2 } { tan }^{ -1 }(x2)$$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 9

${ sin }^{ -1 }\left( \frac { x }{ 5 } \right) +{ cosec }^{ -1 }\left( \frac { 5 }{ 4 } \right) =\frac { \pi }{ 2 }$ , then value of x is :

0%
1
0%
3
0%
4
0%
5

The value of

${ tan }^{ -1 }\left( \frac { xcos\theta }{ 1-xsin\theta } \right) -{ cot }^{ -1 }\left( \frac { cos\theta }{ x-sin\theta } \right)$ is :

0%
$2\theta$
0%
$\theta$
0%
$\frac { \theta }{ 2 }$
0%
Independent of $\theta$

Solve :

$sin h^{-1} (\dfrac{x}{\sqrt{1-x^2}}) =$

0%
$cos h^{-1} x$
0%
$-cos h^{-1} x$
0%
$tan h^{-1} x$
0%
$-tan h^{-1} x$

The number of solution of the equation

${\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$ is are

0%
0
0%
1
0%
2
0%
More then Two

The value of

$\tan^{-1}\left[\dfrac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right],\left|x\right|<\dfrac{1}{2},x=0$ , is equal to:

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

${ sin }^{ -1 }\left( a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } -...........\infty \right) +{ cos }^{ -1 }\left( 1+b+{ b }^{ 2 }+.......\infty \right) =\frac { \pi }{ 2 }$

0%
$a=-3,b=1$
0%
$a=1,b=\dfrac { -1 }{ 3 }$
0%
$a=\dfrac { 1 }{ 6 } ,b=\dfrac { 1 }{ 2 }$
0%
$a=\dfrac { 1 }{ 6 } ,b=\dfrac { 1 }{ 3 }$

$(cot^{-1}x)^{2}-3(cot^{-1}x)+2 > 0,$ then x lies in

0%
(cot 2, cot 1)
0%
$(-\infty , cot 2)\cup (cot 1,\infty )$
0%
$(cot 1 \infty)$
0%
$(-\infty , cot 1)\cup (cot 2,\infty )$

Thee value of

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

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

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

The solution of the equation

$sin^{-1}(\frac{dt}{dx})=x+y$ is

0%
tan(x+y)+sec(x+y)=x+c
0%
tan(x+y)-sec(x+y)=x+c
0%
tan(x+y)+sec(x+y)+x+c=0
0%
None of these

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

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

Let

$S_{n}=\cot^{-1}\left(3x+\dfrac{2}{x}\right)+ \cot^{-1}\left(6x+\dfrac{2}{x}\right)+ \cot^{-1}\left(10x+\dfrac{2}{x}\right)+.....+n$ term where

$x>0$ . If

$\displaystyle \lim_{n\rightarrow \infty}S_{n}=$ then

$x$ equals

0%
$\dfrac{\pi}{4}$
0%
$1$
0%
$\tan 1$
0%
$\cot 1$

Let f:-

$\left\{ 0,4\pi \right\}$

$\rightarrow \left[ 0,\pi \right]$ be defined by f(x)=

${ cos }^{ -1 }\left( cosx \right)$ . The number of points x

$\in \left[ 0,4\pi \right]$ satisfying the equation f(x)=

$\frac { 10-x }{ 10 }$ is

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

Sum of maximum and minimum values of (sin

$^{-1}$ x)

$^4$ + (cos

$^{-1}$ x)

$^4$ is:

0%
$\dfrac{137\pi^2}{128}$
0%
$\dfrac{\pi^2}{17}$
0%
$\dfrac{17\pi^4}{16}$
0%
$\dfrac{137\pi^4}{128}$

Number of values of x for which

$(tan^{-1} x)^2 + (cot^{-1} x)^2 = \dfrac{\pi^2}{4}$ is

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

${ tan }^{ -1 }\frac { \sqrt { 1+{ x }^{ 2 }-1 } }{ x } =4\quad then\quad x$

0%
tan8
0%
tan4
0%
$tan\frac { 1 }{ 4 }$
0%
tan8

$f(x)=2tan^{-1}x+sin^{-1}\left (\dfrac {2x}{1+x^2}\right )$ then for x>1, f(x)=

0%
$sec^{-1}x$
0%
$sin^{-1}x$
0%
$\pi$
0%
$\cfrac {\pi }{2}$

${ \tan \text{h} }^{ -1 }\left( \dfrac { 1 }{ 3 } \right) =\dfrac { 1 }{ 2 } \log t ,$ then

$t$ is equal to

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

$\displaystyle \sum^{\infty}_{r=1}\tan^{-1}\left(\dfrac {1}{r^{2}+5r+7}\right)$ equal to

0%
$\tan^{-1}3$
0%
$\dfrac {3\pi}{4}$
0%
$\sin^{-1}\dfrac {1}{\sqrt {10}}$
0%
$\cot^{-1}2$

$tan^{-1}y=tan^{-1}x+tan^{-1}(\frac{2x}{1-x^{2}})$ where

$|x| < \frac{1}{\sqrt{3}}$ . Then a value of y is:

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

${ tanh }^{ -1 }\left( \frac { 1 }{ 3 } \right) +{ coth }^{ -1 }\left( 3 \right) =.....$

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

The value of sin

$\left ( 3 sin^{-1}\left ( 0.8 \right ) \right )$ is

0%
sin(2)
0%
sin(1.88)
0%
-sin(0.88)
0%
None of these

The value of

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

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$\pi$
0%
None

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

0%
$f\left( \dfrac { 2 }{ 3 } \right) =\dfrac { \pi }{ 3 }$
0%
$f\left( \dfrac { 2 }{ 3 } \right) =2\cos ^{ -1 }{ \dfrac { 2 }{ 3 } -\dfrac { \pi }{ 3 } }$
0%
$f\left( \dfrac { 1 }{ 3 } \right) =\dfrac { \pi }{ 3 }$
0%
$f\left( \dfrac { 1 }{ 3 } \right) =2\cos ^{ -1 }{ \dfrac { 1 }{ 3 } -\dfrac { \pi }{ 3 } } m$

$(\cos^{-1}x)^{2}+(\cos^{-1}y)^{2}+2(\cos^{-1}x)(\cos^{-1}y)=4\pi^{2}$ then

$x^{2}+y^{2}$ is equal to

0%
$1$
0%
$\dfrac{3}{2}$
0%
$2$
0%
`Will depends on x and y

$\cos ^{ -1 }{ \left( \cos { \dfrac { 7\pi }{ 6 } } \right) }$ is equal to

0%
$\dfrac {7\pi}{6}$
0%
$\dfrac {5\pi}{6}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{6}$

The value of

$\sin^{-1}(\sin 3)+\cos^{-1}(\cos 7)-\tan^{-1}(\tan 5)$ is

0%
$\pi-1$
0%
$\pi$
0%
$3\pi-1$
0%
$2\pi-1$

$\sum^ { \infty }_ { n=1 } tan^{-1}\dfrac{4n}{n^4-2n^2+2}$ is equal to:

0%
$tan^{-1}2+tan^{-1}3$
0%
$4 tan^{-1}1$
0%
$\pi/2$
0%
$sec^{-1}(-\sqrt2)$

The smallest and largest value of

$\tan^{-1}\left(\dfrac {1-x}{1+x}\right),0 \le x \le 1$ are

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

${ cos }^{ -1 }(\frac { x }{ 3 } )+{ cos }^{ -1 }(\frac { y }{ 2 } )=(\frac { \theta }{ 2 } )$ , then the value of

${ 4x }^{ 2 }$ -12xy cos

$(\frac { \theta }{ 2 } )$ +

${ 9y }^{ 2 }$ is equal to

0%
$18(1+cos\theta )$
0%
$18(1-cos\theta )$
0%
$36(1+cos\theta )$
0%
$36(1-cos\theta )$

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

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

$x=si{ n }^{ -1 }(sin10)$ and

$y={ s }^{ -1 }(cos10)$ then

$y-x$ is equal to:

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

Evaluate

$\cot ^{ -1 }{ \sum _{ n=1 }^{ 19 }{ \cot ^{ -1 }{ [1+\sum _{ p=1 }^{ n }{ 2p } ] } } }$

0%
$\frac { 23 }{ 22 }$
0%
$\frac { 19 }{ 23 }$
0%
$\frac { 23 }{ 19 }$
0%
$\frac { 22 }{ 23 }$

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

$y={ cos }^{ -1 }(cos10)$ , then the value of (y - x) is

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

Considering only the principal values of inverse functions, the set

$A=\left\{ x\quad \ge \quad 0\quad :tan^{ -1 }(2x)+tan^{ -1 }(3x)=\dfrac { \pi }{ 4 } \right\}$

0%
Is an empty set
0%
contains more than two elements
0%
contains two elements
0%
is a singleton

The value of

$\sin ^{ -1 }{ (\cos { (\cos ^{ -1 }{ (\cos { x } ) } +\sin ^{ -1 }{ (\sin { x } ) } ) } ) } ,\quad where\quad x\in (\frac { \pi }{ 2 } ,\pi )$ , is equal to

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

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

$sin\theta =$

0%
$tan\cfrac { 1 }{ 2 } x$
0%
${ tan }^{ 2 }(x/2)$
0%
$\cfrac { 1 }{ 2 } { tan }^{ -1 }(x2)$
0%
None of these

The value of

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

0%
$0$
0%
$\pi$
0%
$8\pi +24$
0%
$8\pi -24$

Explanation

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

$={\cos}^{-1}{\sqrt{1-{\sin}^{2}{12}}}-{\cos}^{-1}{\left(\cos{12}\right)}$ since

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

$={\cos}^{-1}{\left(\cos{12}\right)}-{\cos}^{-1}{\left(\cos{12}\right)}=0$

$\tan ^{ -1 }{ (\frac { 5 }{ 12 } ) } +\sin ^{ -1 }{ (\frac { 24 }{ 25 } ) } =\cos ^{ -1 }{ (x) } \Rightarrow x=$

0%
$\frac { -31 }{ 325 }$
0%
$\frac { -33 }{ 325 }$
0%
$\frac { -36 }{ 325 }$
0%
$\frac { -39 }{ 325 }$

$\cot \dfrac { 2{ x } }{ 3 } +\tan \dfrac { { x } }{ 3 } =\csc \dfrac { { kx } }{ 3 } ,$ then the value of

$\tan ^{ { -1 } } (\tan { k } )$ equals

0%
$2$
0%
$2 - \pi$
0%
$\pi - 2$
0%
$2\pi - 2$

$\sin ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)+\sin ^{-1}\left(\dfrac{\sqrt{2}-1}{\sqrt{6}}\right)+\ldots+\sin ^{-1}\left(\dfrac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)+\ldots . \infty=$

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

$4\tan ^{ -1 }{ \frac { 1 }{ 5 } } -\tan ^{ -1 }{ \frac { 1 }{ 70 } } +\tan ^{ -1 }{ \frac { 1 }{ 99 } } =$

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

The value of the expression tan

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

0%
$2-\sqrt{5}$
0%
$\sqrt{5}-2$
0%
$\frac{\sqrt{5}-2}{2}$
0%
5- $\sqrt{2}$

$x = \sin ^ { - 1 } ( \sin 10 ) \text { and } y = \cos ^ { - 1 } ( \cos 10 )$ then

$y - x$ is equal to:

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

$cos{ }^{ -1 }\frac { x }{ a } +cos{ }^{ -1 }\frac { y }{ b } =\alpha \quad then\quad \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { 2xy }{ ab } cos\alpha +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =$

0%
$sin^{2}\alpha$
0%
$cos^{2}\alpha$
0%
$tan^{2}\alpha$
0%
$cot^{2}\alpha$

What is the value of

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

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

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

$\sin x$ is equal to

0%
$\displaystyle {\tan}^{2}\frac{\alpha}{2}$
0%
$\displaystyle {\cot}^{2}\frac{\alpha}{2}$
0%
$\tan\alpha$
0%
$\displaystyle \cot\frac{\alpha}{2}$

The product of all values of x satisfying the equation.

${ sin }^{ -1 }cos\left( \dfrac { { 2x }^{ 2 }+10\left| x \right| +4 }{ { x }^{ 2 }+5\left| x \right| +3 } \right)$

$=cot\left\{ { cot }^{ -1 }\left( \dfrac { 2-18\left| x \right| }{ 9\left| x \right| } \right) \right\} +\dfrac { \pi }{ 2 }$ is

0%
9
0%
-9
0%
3
0%
-1

The value of

$\displaystyle sin^{1}\left ( sin\dfrac{5\pi}{3} \right )$ is ......

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

The value of

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

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

$( x \in ( 0,1 ) )$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
None

$f ( x ) = \sin ^ { - 1 } \sqrt { \frac { \sqrt { 1 + x ^ { 2 } } - 1 } { 2 \sqrt { 1 + x ^ { 2 } } } } ,$ then which of the following is (are) correct?

0%
$f ( -1 ) = \frac { - 1 } { 4 }$
0%
$\operatorname { Ran } \mathscr { f } _ { f ( x ) }$ is $\left[ 0 , \frac { \pi } { 2 } \right]$
0%
$f ^ { \prime } ( x )$ is an odd function
0%
$\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x } = \frac { 1 } { 2 }$

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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