If $${ \cot }^{ -1 }(\sqrt { \cos a } )-{ \tan }^{ -1 }(\sqrt { \cos a } )=x,$$ then $$\sin x$$ is equal to

$${\tan }^{ 2 }\dfrac { a }{ 2 } \quad $$

$${ \cot }^{ 2 }\dfrac { a }{ 2 } $$

If$${\cos ^{ - 1}}\left( {2{x^2} - 1} \right) = 2\pi - 2{\cos ^{ - 1}}x,$$ then

$$x \in \left[ { - 1,0} \right]$$

$$x \in \left[ {0,\frac{1}{{\sqrt 2 }}} \right]$$

$$x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$$

If $$\cos^{-1} x > \sin^{-1} x$$, then find the range of x

$$-1 \leq x < \dfrac {1}{\sqrt {2}}$$

$$0\leq < \dfrac {1}{\sqrt {2}}$$

$$\dfrac {1}{\sqrt {2}} < x \leq 1$$

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 11

The value of

t a n^{2} (s e c^{- 1} 3) + c o t^{2} (c o s e c^{- 1} 4)

$tan^{2}(sec^{-1}3)+cot^{2}(cosec^{-1}4)$ is -

0%
9
0%
16
0%
25
0%
23

{c o s}^{- 1} [2 c o t^{- 1} (\sqrt 2 - 1)]

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

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

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

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

sin x

$\sin x$ is equal to

0%
${ \cot }\dfrac { a }{ 2 }$
0%
${ \cot }^{ 2 }\dfrac { a }{ 2 }$
0%
${\tan }^{ 2 }\dfrac { a }{ 2 } \quad$
0%
$\tan a$

{t a n}^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 + {- x}^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 {- x}^{2}}}), [x] \leq \frac{1}{\sqrt{2}}, i s e q u a l t o

${ tan }^{ -1 }\left( \dfrac { \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1+{ -x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1{ -x }^{ 2 } } } \right) ,[x]\le \dfrac { 1 }{ \sqrt { 2 } } ,is\quad equal\quad to$

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

s i n^{- 1} x + s i n^{- 1} y = \frac{2 π}{3}

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

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

t a n^{- 1} (1 - x^{2} - \frac{1}{x^{2}}) + s i n^{- 1} (x^{2} + \frac{1}{x^{2}} - 1)

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

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

c o t^{1} \frac{1}{x} + c o t^{- 1} \frac{1}{y} + c o t^{- 1} \frac{1}{z} = \frac{π}{2}

$cot^{1}\frac{1}{x}+cot^{-1}\frac{1}{y}+cot^{-1 }\frac{1}{z}=\frac{\pi }{2}$ then.....

0%
x+y+z=xyz
0%
xy+yz+zx=1
0%
x+y+z=1
0%
$x^{2}+y^{2}+z^{2}=1$

{cos}^{- 1} (2 x^{2} - 1) = 2 π - 2 {cos}^{- 1} x,

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

0%
$x \in \left[ { - 1,0} \right]$
0%
$x \in \left[ {0,1} \right]$
0%
$x \in \left[ {0,\frac{1}{{\sqrt 2 }}} \right]$
0%
$x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]$

4 s i n^{- 1} x + c o s^{- 1} x = π

$4 sin^{-1}x+cos^{-1} x=\pi$ then x=

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

Evaluate :

sin (\frac{1}{2} {cos}^{- 1} \frac{4}{5})

$\sin \left(\dfrac {1}{2}\cos^{-1}\dfrac {4}{5}\right)$

0%
$\dfrac {1}{\sqrt 5}$
0%
$\dfrac {2}{\sqrt 5}$
0%
$\dfrac {1}{\sqrt {10}}$
0%
$\dfrac {2}{\sqrt {10}}$

α = 2 {tan}^{- 1} (\sqrt{3 - 2 \sqrt{2}}) + {sin}^{- 1} (\frac{1}{\sqrt{6} - \sqrt{2}}), β = {cot}^{- 1} (\sqrt{3} - 2) + \frac{1}{8} {sec}^{- 1} (- 2)

$\alpha =2\tan^{-1}(\sqrt{3-2\sqrt{2}})+\sin^{-1}\left(\dfrac{1}{\sqrt{6}-\sqrt{2}}\right), \beta =\cot^{-1}(\sqrt{3}-2)+\dfrac{1}{8}\sec^{-1}(-2)$ &

γ = {tan}^{- 1} \frac{1}{\sqrt{2}} + {cos}^{- 1} \frac{1}{\sqrt{3}}

$\gamma =\tan^{-1}\dfrac{1}{\sqrt{2}}+\cos^{-1}\dfrac{1}{\sqrt{3}}$ , then?

0%
$\alpha =\beta$
0%
$\alpha +\beta =3\gamma$
0%
$4(\beta -\gamma)=\alpha$
0%
$\beta =\gamma$

{cos}^{- 1} x > {sin}^{- 1} x

$\cos^{-1} x > \sin^{-1} x$ , then find the range of x

0%

$-1 \leq x < \dfrac {1}{\sqrt {2}}$
0%
$0\leq < \dfrac {1}{\sqrt {2}}$
0%
$\dfrac {1}{\sqrt {2}} < x \leq 1$
0%
$x > 0$

Explanation

We have

{cos}^{- 1} x > {sin}^{- 1} x

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

\Rightarrow \frac{π}{2} - {sin}^{- 1} x > {sin}^{- 1} x

$\Rightarrow\,\dfrac{\pi}{2}-{\sin}^{-1}{x}>{\sin}^{-1}{x}$

\Rightarrow \frac{π}{2} > 2 {sin}^{- 1} x

$\Rightarrow\,\dfrac{\pi}{2}>2{\sin}^{-1}{x}$

\Rightarrow \frac{π}{4} > {sin}^{- 1} x

$\Rightarrow\,\dfrac{\pi}{4}>{\sin}^{-1}{x}$

\Rightarrow {sin}^{- 1} x < \frac{π}{4}

$\Rightarrow\,{\sin}^{-1}{x}<\dfrac{\pi}{4}$ .......

(1)

$(1)$

But

- \frac{π}{2} \leq {sin}^{- 1} x < \frac{π}{2}

$-\dfrac{\pi}{2}\le {\sin}^{-1}{x}<\dfrac{\pi}{2}$ ........

(2)

$(2)$

From

(1)

$(1)$ and

(2)

$(2)$ we have

- \frac{π}{2} \leq {sin}^{- 1} x < \frac{π}{4}

$-\dfrac{\pi}{2}\le {\sin}^{-1}{x}<\dfrac{\pi}{4}$

\Rightarrow sin (- \frac{π}{2}) \leq x < sin \frac{π}{4}

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

\Rightarrow - 1 \leq x < \frac{1}{\sqrt{2}}

$\Rightarrow\,-1\le x<\dfrac{1}{\sqrt{2}}$

{cot}^{- 1} (\frac{- 1}{5}) = x

$\cot^{-1}\left(\dfrac {-1}{5}\right)=x$ then

sin x =

$\sin x=$ ?

0%
$\dfrac {1}{\sqrt {26}}$
0%
$\dfrac {5}{\sqrt {26}}$
0%
$\dfrac {1}{\sqrt {24}}$
0%
$none\ of\ these$

The value of

{tan}^{- 1} (\frac{x cos θ}{1 - x sin θ}) - {cot}^{- 1} (\frac{cos θ}{x - sin θ})

$\tan^{-1}\Bigg(\dfrac{x\cos\theta}{1-x\sin\theta}\Bigg)-\cot^{-1}\Bigg(\dfrac{\cos\theta}{x-\sin\theta}\Bigg)$ is

0%
$2\theta$
0%
$\theta$
0%
$\dfrac{\theta}{2}$
0%
independent of $\theta$

t a n^{- 1} (\frac{c_{1} x - y}{c_{1} y + x}) + t a n^{- 1} (\frac{c_{2} - c_{1}}{1 + c_{2} c_{1}}) + t a n^{- 1} (\frac{c_{3} - c_{2}}{1 + c_{3} c_{2}}) +

$tan^{-1} \left (\dfrac{c_1\,x - y}{c_1\,y + x} \right ) + tan^{-1} \left ( \dfrac{c_2 - c_1}{1 + c_2c_1} \right ) + tan^{-1} \left (\dfrac{c_3 - c_2}{1 + c_3c_2} \right ) +$ .....

+ t a n^{- 1} (\frac{1}{c_{n}}) =

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

0%
$tan^{-1} \,(y/x)$
0%
$tan^{-1}\dfrac{x}{y}$
0%
$-tan^{-1}\left ( \dfrac{x}{y} \right )$
0%
none of these

Let

$\begin{vmatrix}tan^{-1}x & tan^{-1}2x & tan^{-1}3x\\ tan^{-1}3x & tan^{-1}x & tan^{-1}2x\\ tan^{-1}2x & tan^{-1}3x & tan^{-1}x \end{vmatrix}$ = 0, then the number of values of x satisfying the equation is

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

Choose the correct answer :

$\tan^{-1}\sqrt{3} - \cot^{-1} (-\sqrt{3})$ is equal to

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

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 11 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 11 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.