MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 4

The value of $$\sin^{-1}\left\{ \cot { \left\{ \sin ^{ -1 }{ \sqrt { \frac { 2-\sqrt { 3 } }{ 4 } } +\cos ^{ }{ \frac { \sqrt { 12 } }{ 4 } +\sec ^{ -1 }{ \sqrt { 2 } } } } \right\} } \right\} $$ is equal to

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

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

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

$${\cos ^{ - 1}}\left\{ {\dfrac{1}{{\sqrt 2 }}\left( {\cos \dfrac{{9\pi }}{{10}} - \sin \dfrac{{9\pi }}{{10}}} \right)} \right\} = $$

0%
$$\frac{{23\pi }}{{20}}$$
0%
$$\frac{{7\pi }}{{10}}$$
0%
$$\frac{{7\pi }}{{20}}$$
0%
$$\frac{{17\pi }}{{20}}$$

The solution set of the equation $$\sin^{-1}\sqrt{1-x^2}+\cos^{-1}x=\cot^{-1}\dfrac{\sqrt{1-x^2}}{x}-\sin^{-1}x$$ is?

0%
$$[-1, 1]-\{0\}$$
0%
$$(0, 1] \cup \{-1\}$$
0%
$$[-1, 0)\cup \{1\}$$
0%
$$[-1, 1]$$

Given that $$0 \le x\le \dfrac 12$$ the value of $$\tan \left[ { sin }^{ -1 }\left( \dfrac { x }{ \sqrt { 2 } } +\sqrt { \dfrac { 1-{ x }^{ 2 } }{ 2 } } \right) -{ sin }^{ -1 }x \right]$$ is

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

the value of $$\cos ^{ -2 }{ \left( \cos { 10 } \right) }$$ is

0%
$$4\pi +10$$
0%
$$4\pi -10$$
0%
$$2\pi +10$$
0%
$$2\pi -10$$

$$2\cot ^{ -1 }{ 7 } +\cos ^{ -1 }{ \dfrac { 3 }{ 5 } } $$ is equal to

0%
$$\cot ^{ -1 }{ \left( \dfrac { 44 }{ 117 } \right) } $$
0%
$$cosec ^{ -1 }{ \left( \dfrac { 125 }{ 117 } \right) } $$
0%
$$\tan ^{ -1 }{ \left( \dfrac { 4 }{ 117 } \right) } $$
0%
$$\tan ^{ -1 }{ \left( \dfrac { 44 }{ 125 } \right) } $$

The value of $$\tan ^{ -1 }{ \left( 2\sin { \left( \sec ^{ -1 }{ \left( 2 \right) } \right) } \right) } $$ is

0%
$$\dfrac {\pi}{6}$$
0%
$$\dfrac {\pi}{4}$$
0%
$$\dfrac {\pi}{3}$$
0%
$$\dfrac {\pi}{2}$$

$${\tan ^{ - 1}}\left( {\frac{1}{2}\tan 2A} \right) + {\tan ^{ - 1}}\cot A + {\tan ^{ - 1}}{\cot ^3}A = $$

0%
$$0;\,if\,\pi /4 < A < \pi /2$$
0%
$$0;\,if\,0 < A < \pi /4$$
0%
$$\pi; \,if\pi /4 < A < \pi /2$$
0%
$$\pi ;\,if0 < A < \pi /2$$

Explanation

We know that $${\tan}^{-1}{x}+{\tan}^{-1}{y}=\begin{cases} {\tan}^{-1}{\left(\dfrac{x+y}{1-xy}\right)};if\,xy<1 \\ \pi+{\tan}^{-1}{\left(\dfrac{x+y}{1-xy}\right)};if\,xy>1 \end{cases}$$

Now, for $$0<A<\dfrac{\pi}{4}$$ we have $$\cot{A}>1$$

and for $$\dfrac{\pi}{4}<A<\dfrac{\pi}{2}$$ we have $$\cot{A}<1$$

$$\therefore\,$$ we can write

$${\tan}^{-1}{\left(\cot{A}\right)}+{\tan}^{-1}{\left({\cot}^{3}{A}\right)}=\begin{cases} {\tan}^{-1}{\left(\dfrac{\cot{A}+{\cot}^{3}{A}}{1-{\cot}^{4}{A}}\right)};if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi+{\tan}^{-1}{\left(\dfrac{\cot{A}+{\cot}^{3}{A}}{1-{\cot}^{4}{A}}\right)};if\,0<A<\dfrac{\pi}{2} \end{cases}$$

$$=\begin{cases} {\tan}^{-1}{\left(\dfrac{\cot{A}\left(1+{\cot}^{2}{A}\right)}{\left(1-{\cot}^{2}{A}\right)\left(1+{\cot}^{2}{A}\right)}\right)};if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi+{\tan}^{-1}{\left(\dfrac{\cot{A}\left(1+{\cot}^{2}{A}\right)}{\left(1-{\cot}^{2}{A}\right)\left(1+{\cot}^{2}{A}\right)}\right)};if\,0<A<\dfrac{\pi}{2} \end{cases}$$

$$=\begin{cases} {\tan}^{-1}{\left(\dfrac{\cot{A}}{1-{\cot}^{2}{A}}\right)};if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi+{\tan}^{-1}{\left(\dfrac{\cot{A}}{1-{\cot}^{2}{A}}\right)};if\,0<A<\dfrac{\pi}{2} \end{cases}$$

$$=\begin{cases} {\tan}^{-1}{\left(\dfrac{\dfrac{1}{\tan{A}}}{1-\dfrac{1}{{\tan}^{2}{A}}}\right)};if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi+{\tan}^{-1}{\left(\dfrac{\dfrac{1}{\tan{A}}}{1-\dfrac{1}{{\tan}^{2}{A}}}\right)};if\,0<A<\dfrac{\pi}{2} \end{cases}$$

$$=\begin{cases} {\tan}^{-1}{\left(\dfrac{-\tan{A}}{1-{\tan}^{2}{A}}\right)};if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi+{\tan}^{-1}{\left(\dfrac{-\tan{A}}{1-{\tan}^{2}{A}}\right)};if\,0<A<\dfrac{\pi}{2} \end{cases}$$

$$=\begin{cases} {\tan}^{-1}{\left(-\dfrac{1}{2}\tan{2A}\right)};if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi+{\tan}^{-1}{\left(-\dfrac{1}{2}\tan{2A}\right)};if\,0<A<\dfrac{\pi}{2} \end{cases}$$

$$=\begin{cases} -{\tan}^{-1}{\left(\dfrac{1}{2}\tan{2A}\right)};if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi-{\tan}^{-1}{\left(\dfrac{1}{2}\tan{2A}\right)};if\,0<A<\dfrac{\pi}{2} \end{cases}$$

Adding both sides $${\tan}^{-1}{\left(\dfrac{1}{2}\tan{2A}\right)}$$ on both sides, we get

$${\tan}^{-1}{\left(\dfrac{1}{2}\tan{2A}\right)}+{\tan}^{-1}{\left(\cot{A}\right)}+{\tan}^{-1}{\left({\cot}^{3}{A}\right)}=\begin{cases} 0;if\,\dfrac{\pi}{4}<A<\dfrac{\pi}{2} \\ \pi;if\,0<A<\dfrac{\pi}{2} \end{cases}$$

The value of $$\cos ^{ -1 }{ \left\{ \frac { \sqrt { 1-\sin { x } } +\sqrt { 1+\sin { x } } }{ \surd \left( 1-\sin { x } \right) -\surd \left( 1+\sin { x } \right) } \right\} }$$ is $$\left( 0<x<2\pi \right)$$

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

Annual expenses of A and B are in the ratio $$5:3$$. The saving of Aand B are in the ratio $$1:2$$. Find the expenses of A. given that the income of A is $$Rs. 8000$$ and that of B is $$Rs.9000$$.

0%
$$3000$$
0%
$$4000$$
0%
$$4500$$
0%
$$5000$$

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

0%
$${ tan }^{ -1 }\left( \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right)$$
0%
$$2\tan ^{ -1 }{ \left(- \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right) }$$
0%
$$\dfrac { 1 }{ 2 } \tan ^{ -1 }{ \left( 2\tan { \dfrac { x }{ 2 } } \right) }$$
0%
$$2\tan ^{ -1 }{ \left( \dfrac { 1 }{ 2 } \tan { \dfrac { x }{ 2 } } \right) }$$

Explanation

Let $$\cos^{-1}\dfrac{(3+5\cos x)}{(5+3\cos x)}=\alpha$$

$$\dfrac{3+5\cos x}{5+3\cos x}=\cos \alpha$$

$$\because \cos \alpha\dfrac{[1-\tan^2(\dfrac{\alpha}{2})]}{[1+\tan^2(\dfrac{x}{2})]}$$

$$\cos \alpha=5\dfrac{\dfrac{[1-\tan^2(\dfrac{x}{2})]}{[1+\tan^2(\dfrac{x}{2})]}+3}{5+3[\dfrac{1-\tan^2(\dfrac{x}{2})}{1+\tan^2(\dfrac{x}{2})}]}$$

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

$$\Rightarrow \dfrac{5-5\tan^2(\dfrac{x}{2})+3+3\tan^2(\dfrac{x}{2})}{5+5\tan^2(\dfrac{x}{2})+3-3\tan^2(\dfrac{x}{2})}$$

$$\Rightarrow \dfrac{8-2\tan^2(\dfrac{x}{2})}{8+2\tan^2(\dfrac{x}{2})}$$

$$\dfrac{4-\tan^2(\dfrac{x}{2})}{4+\tan^2(\dfrac{x}{2})}$$

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

$$\Rightarrow \sin \alpha=\sqrt{1-\left(\dfrac{4-\tan^2\left(\dfrac{x}{2}\right)}{4+\tan^2\left(\dfrac{x}{2}\right)}\right)^2}$$

$$\Rightarrow \sqrt{\dfrac{\left[4+\tan^2(\dfrac{x}{2}\right)^2]\left[4+\tan^2(\dfrac{x}{2})\right]}{4+\tan^2\left(\dfrac{x}{2}\right)}^2}^2$$

$$=\sqrt{\dfrac{4+\tan^2\left(\dfrac{x}{2}\right)-4+\tan^2\left(\dfrac{x}{2}\right)\left(4+\tan^2(\dfrac{x}{2}+4-\tan^2\right)\left(\dfrac{x}{2})\right)}{\left[4+\tan^2(\dfrac{x}{2})\right]^2}}$$

$$\Rightarrow \dfrac{\sqrt{2\tan^2(\dfrac{x}{2}).8}}{4+\tan^2(\dfrac{x}{2})}\Rightarrow \dfrac{4\tan(\dfrac{x}{2})}{4\tan^2(\dfrac{n}{2})}$$

$$\because \tan\left(\dfrac{x}{2}\right)=\dfrac{1-\cos \alpha}{\sin \alpha}$$

$$\therefore \dfrac{1-\left(\dfrac{4-\tan^2(\dfrac{n}{2})}{4+\tan^2(\dfrac{n}{2})}\right)}{\left(\dfrac{4\tan(\dfrac{n}{2})}{4+\tan^2(\dfrac{n}{2})}\right)}\Rightarrow \dfrac{4+\tan^2\left(\dfrac{n}{2}-4+\tan^2\right)\left(\dfrac{n}{2}\right)}{4\tan \left(\dfrac{n}{2}\right)}$$

$$\tan\left(\dfrac{x}{2}\right)\Rightarrow \dfrac{2\tan^2\left(\dfrac{x}{2}\right)}{2^4\tan\left(\dfrac{x}{2}\right)}\Rightarrow \dfrac{1}{2}\tan\left(\dfrac{x}{2}\right)$$

$$\dfrac{\alpha}{2}=\tan^{-1}\left[\dfrac{1}{2}\tan(\dfrac{x}{2})\right]$$

$$\boxed{\cos^{-1}\left(\dfrac{5 \cos x+3}{5+3\cos x}\right)\Rightarrow 2\tan^{-1}\left[\dfrac{1}{2}\tan(\dfrac{n}{2})\right]}$$

So, option (b) is correct.

If $$\sin^{-1}\left(x-\dfrac {x^{2}}{2}+\dfrac {x^{3}}{4}-...\infty \right)+\cos^{-1}\left(x^{2}-\dfrac {x^{4}}{2}+\dfrac {x^{6}}{4}-...\infty \right)=\dfrac {\pi}{2}$$ for $$0 < |x| < \sqrt {2}$$,then $$x$$ equal

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

$${ cos }^{ -1 }\left( \dfrac { \cos { \alpha } \cos { \beta } }{ 1+\cos { \alpha } \cos { \beta } } \right) =2{ tan }^{ -1 }\left( \tan { \dfrac { \alpha }{ 2 } } \tan { \dfrac { \beta }{ 2 } } \right)$$.

0%
True
0%
False

If $${sin}^{-1}x+{sin}^{-1}y+{sin}^{-1}z=\pi$$, then $$x\sqrt {{1}-{x}^{2}}+y\sqrt {{1}-{y}^{2}}+z \sqrt {{1}-{z}^{2}}=-2xyz$$

0%
True
0%
False

State true or false.
$${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{8}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) = \dfrac{\pi }{2}$$

0%
True
0%
False

The value of $$\cos\left\{\tan^{-1}\left(\tan \dfrac{15\pi}{4}\right)\right\}$$ is?

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

If the number $$93215x2$$ is completely divisible by $$11$$, then $$x$$ is equal to

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

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

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

The value of $$3\tan^{-1}\dfrac {1}{2}+2\tan^{-1}\dfrac {1}{5}+\sin^{-1}\dfrac {142}{65\sqrt {5}}$$ is

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

Find the approximate value of $$\sin^{-1}(0.51)$$ given that $$\sqrt {3}=1.7321$$

0%
$$\dfrac {\pi}{6}+0.011541$$
0%
$$\dfrac {\pi}{3}+0.011541$$
0%
$$\dfrac {\pi}{6}+0.011547$$
0%
$$\dfrac {\pi}{3}+0.011547$$

$$\cot ^{ -1 }{ \left( \dfrac { \sqrt { 1-\sin { x } } +\sqrt { 1+\sin { x } } }{ \sqrt { 1-\sin { x } } -\sqrt { 1+\sin { x } } } \right) }$$=....$$\left( 0<x<\dfrac { \pi }{ 2 } \right)$$

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

The value of $$\tan^{-1}\dfrac {1}{2}+\tan^{-1}\dfrac {1}{3}$$ is

0%
$$\dfrac {\pi}{4}$$
0%
$$\dfrac {\pi}{6}$$
0%
$$\dfrac {\pi}{3}$$
0%
$$0$$

Let in $$\Delta ABC, \, \angle A = \dfrac{\pi}{2}$$. Then value of $$\tan^{-1} \dfrac{b}{a + c} + \tan^{-1} \dfrac{c}{a + b}$$ equals

0%
$$\dfrac{\pi}{2}$$
0%
$$\dfrac{\pi}{4}$$
0%
$$\dfrac{\pi}{3}$$
0%
None of these

$${\sin ^{ - 1}}(\cos \left( {{{\sin }^{ - 1}}x} \right)) + {\cos ^{ - 1}}(\sin \left( {{{\cos }^{ - 1}}x} \right)){\text{is}}\;{\text{equal}}\;{\text{to}}{\text{.}}$$

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

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

0%
$$a$$
0%
$${a^{3\,}}$$
0%
$${a^2} - a + 1\,$$
0%
$${a^2} + a - 1$$

$${\cot ^{ - 1}}\left( {2 + \sqrt 3 } \right) = $$

0%
$$\frac{\pi }{{12}}$$
0%
$$\frac{\pi }{{15}}$$
0%
$$\frac{\pi }{5}$$
0%
$$\frac{{3\pi }}{{10}}$$

If minimum value of $${\left( {{{\sin }^{ - 1}}x} \right)^2} + {\left( {{{\cos }^{ - 1}}x} \right)^2} = \frac{{{\pi ^2}}}{K}$$ then the value of $$K$$ is

0%
$$4$$
0%
$$8$$
0%
$$6$$
0%
$$16$$

The value of $${\sin ^{ - 1}}\left( {\cos \dfrac{{33\pi }}{5}} \right)$$ is

0%
$$\dfrac{{3\pi }}{5}$$
0%
$$\dfrac{{7\pi }}{5}$$
0%
$$\dfrac{\pi }{{10}}$$
0%
$$ - \dfrac{\pi }{{10}}$$

The value of $$\cos(\tan^{-1}(\tan2))$$

0%
$$\dfrac{1}{\sqrt{5}}$$
0%
$$-\dfrac{1}{\sqrt{5}}$$
0%
$$\cos 2$$
0%
$$-\cos 2$$

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

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

$${\cos ^{ - 1}}\dfrac{3}{5} - {\sin ^{ - 1}}\dfrac{4}{5} = {\cos ^{ - 1}}x$$ then $$x$$ is equal to:

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

If $$\sin ^{ -1 }{ x } =\cot ^{ -1 }{ x } $$ then

0%
$${ x }^{ 2 }=\cfrac { \sqrt { 5 } -1 }{ 2 } $$
0%
$${ x }^{ 2 }=\cfrac { \sqrt { 5 } +1 }{ 2 } $$
0%
$${ x }^{ }=\cfrac { \sqrt { 5 } +1 }{ 2 } $$
0%
$${ x }^{ }=\cfrac { \sqrt { 5 } -1 }{ 2 } $$

Explanation

$$\mathbf{{\text{Step - 1: Writing co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x in terms of si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$$

$${\text{Let co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x = }}\theta $$

$$ \Rightarrow {\text{ x = cot }}\theta $$

$${\text{We know that, 1 + co}}{{\text{t}}^{\text{2}}}\theta {\text{ = cose}}{{\text{c}}^{\text{2}}}\theta $$

$$\therefore {\text{ 1 + }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{1}{{{\text{si}}{{\text{n}}^{\text{2}}}\theta }}$$

$$ \Rightarrow {\text{ si}}{{\text{n}}^{\text{2}}}\theta {\text{ = }}\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}$$

$$ \Rightarrow {\text{ sin }}\theta {\text{ = }}\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} $$

$$ \Rightarrow {\text{ }}\theta {\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$

$$ \Rightarrow {\text{ co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$

$$\mathbf{{\text{Step - 2: Equating it to si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$$

$${\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$

$$ \Rightarrow {\text{ x = }}\left( {\sqrt {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} } \right)$$

$$ \Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}$$

$$ \Rightarrow {\text{ }}{{\text{x}}^{\text{4}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - 1 = 0}}$$

$$ \Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{{\text{ - 1 }} \pm {\text{ }}\sqrt {{\text{1 + 4}}} }}{{\text{2}}}$$

$$\because {\text{ }}{{\text{x}}^{\text{2}}}{\text{ cannot be negative}}$$

$$ \Rightarrow {\text{ }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{\sqrt {\text{5}} {\text{ - 1}}}}{{\text{2}}}$$

$$\mathbf{{\text{Hence, If si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x = co}}{{\text{t}}^{{\text{ - 1}}}}{\text{x then }}{{\text{x}}^{\text{2}}}{\text{ = }}\dfrac{{\sqrt {\text{5}} {\text{ - 1}}}}{{\text{2}}}}$$

$$Sin^{-1}(2x(1-x^2)) = 2sin^{-1}x$$ is true if.

0%
$$x\in [0,1]$$
0%
[$$-\dfrac{1}{2} ,\dfrac{1}{2}$$]
0%
[$$\dfrac{1}{2},- \dfrac{1}{2}$$]
0%
[$$\dfrac{3}{2}, -\dfrac{3}{2}$$]

The greatest and least value of $${\left( {{{\sin }^{ - 1}}x} \right)^2} + {\left( {{{\cos }^{ - 1}}x} \right)^2}$$ are respectively

0%
$$\dfrac{{{\pi ^2}}}{4}and0$$
0%
$$\dfrac{\pi }{2}and\, - \dfrac{\pi }{2}$$
0%
$$\dfrac{{5{\pi ^2}}}{4}and\dfrac{{{\pi ^2}}}{8}$$
0%
$$\dfrac{{{\pi ^2}}}{4}\,and\,\dfrac{{ - \pi }}{4}$$

The number of elements in the range of
$$f(x)=\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } +\sec ^{ -1 }{ x } $$ is

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

The principal value of $$tan^{-1}[cot\dfrac{3\pi}{4}]$$ is :

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

The algebraic expression for $$\tan \left(\sin^{-1}\cos\ \tan^{-1}\dfrac {x}{2}\right)$$ is

0%
$$\dfrac {2}{x}$$
0%
$$\dfrac {x}{2}$$
0%
$$\dfrac {1}{x}$$
0%
$$\dfrac {2}{|x|}$$

If $$\sin ^{-1}x + \sin ^{-1}y + \sin ^{-1}z = \dfrac{3\pi}{2}$$, then $$(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})$$ is

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

The value of $$\sin \left(\dfrac {1}{4}\sin^{-1}\dfrac {\sqrt {63}}{8}\right)$$ is

0%
$$\dfrac {1}{2}$$
0%
$$\dfrac {1}{3}$$
0%
$$\dfrac {1}{2\sqrt {2}}$$
0%
$$\dfrac {1}{5}$$

If $$\tan^{-1}\left(\dfrac{x+1}{x-1}\right)+\tan^{-1}\left(\dfrac{x-1}{x}\right)=\pi +\tan^{-1}(-7)$$, then $$x=$$

0%
$$2$$
0%
$$-2$$
0%
$$1$$
0%
No solution

The number of real solutions of $$\cos ^ { - 1 } x + \cos ^ { - 1 } 2 x = - \pi$$ is

0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
None

The value of $$ \sin (\cot^{-1}x)$$ is

0%
$$\sqrt{1+x^2}$$
0%
$$x$$
0%
$$(1+x^2)^{-3/2}$$
0%
$$(1+x^2)^{-1/2}$$

$$\cos ^ { - 1 } \left( \cos \left( \dfrac { - 17 \pi } { 5 } \right) \right)$$ is equal to

0%
$$- \dfrac { 17 \pi } { 5 }$$
0%
$$\dfrac { 3 \pi } { 5 }$$
0%
$$\dfrac { 2 \pi } { 5 }$$
0%
none of these

$$\sinh { \left( \cosh ^{ -1 }{ x } \right) } =$$

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

$$\sin\ h^{-1}{\left(2^{3/2}\right)}$$=

0%
$$\log \left(2+\sqrt{8}\right)$$
0%
$$\log \left(3+\sqrt{8}\right)$$
0%
$$\log \left(2-\sqrt{8}\right)$$
0%
$$\log \left(\sqrt{8}+\sqrt{7}\right)$$

The value of $$\tan ^{-1} \left (\dfrac{1}{3}\right) +\tan ^{-1} \left (\dfrac{2}{9}\right) + \tan ^{-1}\left (\dfrac{4}{33}\right) + \tan ^{-1} \left (\dfrac{8}{129}\right) +$$.......$$n$$ terms is

0%
$$\tan^{-1} 2^n - \dfrac{\pi}{4}$$
0%
$$\tan^{-1} 2^n$$
0%
$$\cot^{-1} 2^n$$
0%
$$\dfrac{\sin^{-1} 2^n}{\cos^{-1} 2^n}$$

If $$\cot^{-1} [\sqrt{\cos \alpha}] - \tan^{-1} [\sqrt{\cos \alpha}] = x$$, then $$\sin x$$ is equal to

0%
$$\tan^2 \left (\dfrac{\alpha}{2}\right)$$
0%
$$\cot^2 \left (\dfrac{\alpha}{2}\right)$$
0%
$$\tan \alpha$$
0%
$$\cot \dfrac{\alpha}{2}$$

$$\sec\ h^{-1}\left(\dfrac{1}{5}\right)$$=

0%
$$\log( {\sqrt{24}+5})$$
0%
$$\log {5+\sqrt{27}}$$
0%
$$\log {26+\sqrt{5}}$$
0%
$$\log {27+\sqrt{5}}$$

If $$\cot \dfrac {2x}{3}+\tan \dfrac {x}{3}=\csc \dfrac {x}{3}$$ then value of $$\tan^{-1(\tan k)}$$ equals

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

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 4 - 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 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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