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

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

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9
0%
16
0%
25
0%
23

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

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

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

$\sin x$ is equal to

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

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

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

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

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

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

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

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

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

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

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

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

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

Evaluate :

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

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

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

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

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

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

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

Explanation

We have

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

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

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

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

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

$(1)$

But

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

$(2)$

From

$(1)$ and

$(2)$ we have

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

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

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

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

$\sin x=$ ?

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

The value of

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

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

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

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

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$tan^{-1} \,(y/x)$
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$tan^{-1}\dfrac{x}{y}$
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$-tan^{-1}\left ( \dfrac{x}{y} \right )$
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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

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$1$
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$2$
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$3$
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$4$

Choose the correct answer :

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

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

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

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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