MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 1

Statement I : The equation $$(sin^{-1}x)^3+(cos^{-1}x)^3-a\pi^3=0$$ has a solution for all $$a\geqslant \dfrac {1}{32}.$$
Statement II : For any $$x\epsilon R, sin^{-1}x+cos^{-1}x=\dfrac {\pi}{2}$$ and $$0\leq (sin^{-1}x-\dfrac {\pi}{4})^2\leq \dfrac {9\pi^2}{16}$$.

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Both statements I and II are true.
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Both statements I and II are true but I is not an explanation of II
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Statement I is true and statement II is false
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Statement I is false and statement II is true.

If $$\cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } =\alpha $$, then $$4{ x }^{ 2 }-4xy\cos { \alpha } +{ y }^{ 2 }\quad $$ is equal to

0%
$$2\sin { 2\alpha } $$
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$$4$$
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$$4\sin ^{ 2 }{ \alpha } $$
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$$-4\sin ^{ 2 }{ \alpha } $$

If $$\sin^{-1}\left (\dfrac {2a}{1 + a^{2}}\right ) - \cos^{1}\left (\dfrac {1 - b^{2}}{1 + b^{2}}\right ) = \tan^{-1}\left (\dfrac {2x}{1 - x^{2}}\right )$$, then what is the value of $$x$$?

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$$\dfrac {a}{ b}$$
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$$ab$$
0%
$$\dfrac {b}{a}$$
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$$\dfrac {a - b}{1 + ab}$$

The value of $$\tan^{-1}(1)+\cos^{-1}\left(\dfrac{-1}{2}\right)+\sin^{-1}\left(\dfrac{-1}{2}\right)$$ is equal to

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$$\dfrac{\pi}{4}$$
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$$\dfrac{5\pi}{12}$$
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$$\dfrac{3\pi}{4}$$
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$$\dfrac{13\pi}{12}$$

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

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$$|x|\leq 1$$
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$$x\epsilon R$$
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$$0\leq x\leq 1$$
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$$-1\leq x\leq 0$$

Solve: $$\displaystyle { \tan }^{ -1 }\left( \frac { x }{ y } \right) -{ \tan }^{ -1 }\frac { x-y }{ x+y } $$ is equal to

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

Value of $$\tan^{-1}\begin{Bmatrix}\dfrac{\sin\,2-1}{\cos\,2}\end{Bmatrix}$$ is

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$$\dfrac{\pi}{2}-1$$
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$$1-\dfrac{\pi}{4}$$
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$$2-\dfrac{\pi}{2}$$
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$$\dfrac{\pi}{4}-1$$

The number of triplets $$\left ( x,y,z \right )$$ satisfies the equation $$\displaystyle f\left ( x,y,z \right )= \sin ^{-1}x+\sin ^{-1}y+\sin ^{-1}z= \dfrac{3\pi }{2}$$ is

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$$1$$
0%
$$2$$
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$$0$$
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Infinite

Sin-1 $$(\displaystyle \frac{3}{5})+\mathrm{S}\mathrm{i}\mathrm{n}^{-1}(\frac{5}{13})=\sin_{X}^{-1}$$ then$$\mathrm{x}=$$

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$$(^{\frac{63}{65})}$$
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$$\displaystyle \frac{56}{65}$$
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$$\displaystyle \frac{56}{63}$$
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$$\displaystyle \frac{16}{63}$$

If $$\sec^{-1} x+ \sec^{-1}y + \sec^{-1}z = 3\pi$$, then $$xy + yz + zx =$$ _______.

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

Range of $$f(x)=\tan^{-1}\Bigg[\dfrac{2}{\pi}(2\tan^{-1}x-\sin^{-1}x+\cot^{-1}x-\cos^{-1}x)\Bigg]$$ contains

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Only one integer
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More than 2 integers
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Only two integers
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No integer

The value of $$\sin^{-1} \left( \dfrac{2 \sqrt 2}{3} \right ) + \sin^{-1} \left( \dfrac{1}{3}\right )$$ is equal to

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

$${ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\dfrac { x+y }{ 1-xy } $$, $$xy<1$$

$$=\pi +{ tan }^{ -1 }\dfrac { x+y }{ 1-xy } $$, $$xy>1$$.

Evaluate: $${ tan }^{ -1 }\dfrac { 3sin2\alpha }{ 5+3cos2\alpha } +{ tan }^{ -1 }\left( \dfrac { tan\alpha }{ 4 } \right) $$

where $$-\dfrac { \pi }{ 2 } <\alpha <\dfrac { \pi }{ 2 } $$

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$$\alpha$$
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$$2\alpha$$
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$$3\alpha$$
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$$4\alpha$$

Explanation

(a) $$sin2\alpha =\dfrac { 2t }{ 1+{ t }^{ 2 } } ,cos2\alpha =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } $$, where $$t=tan\alpha $$

L.H.S.$$={ tan }^{ -1 }\dfrac { 3t }{ 4+{ t }^{ 2 } } +{ tan }^{ -1 }\dfrac { t }{ 4 } $$

$$={ tan }^{ -1 }t\dfrac { (16+{ t }^{ 2 }) }{ (16+{ t }^{ 2 }) } ={ tan }^{ -1 }(tan\alpha )=\alpha $$

(b) Let us put $$tan\theta =t$$,

$$2\theta =\dfrac { 2t }{ 1+{ t }^{ 2 } } ,cos2\theta =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } $$.

The 2nd term is

$$\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 3.2t }{ 5(1+{ t }^{ 2 })+(1+{ t }^{ 2 }) } =\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 6t }{ 9+{ t }^{ 2 } } $$

$$=\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 2.t/3 }{ 1+{ \left( t/3 \right) }^{ 2 } } =\dfrac { 1 }{ 2 } { sin }^{ -1 }\dfrac { 2T }{ 1+{ T }^{ 2 } } $$

$$=\dfrac { 1 }{ 2 } .2{ tan }^{ -1 }T={ tan }^{ -1 }(t/3)$$

$$\therefore $$ $$\theta ={ tan }^{ -1 }2{ t }^{ 2 }-{ tan }^{ -1 }\dfrac { t }{ 3 } ={ tan }^{ -1 }\dfrac { 2{ t }^{ 2 }-t/3 }{ 1+2{ t }^{ 2 }.t/3 } $$

or $$tan\theta =\dfrac { t(6t-1) }{ 3+2{ t }^{ 3 } } $$ or $$t(3+2{ t }^{ 3 })=t(6t-1)$$

$$\therefore $$ $$t=0$$ or $$2{ t }^{ 3 }-6t+4=0$$

or $${ t }^{ 3 }-3t+2=0$$

$$\therefore $$ $$t=0$$ or $$(t-1)({ t }^{ 2 }+t-2)=0$$

or $$t=0$$ or $$(t-1)(t-1)(t+2)=0$$

$$\therefore $$ $$t=0,1,-2$$

$$\therefore $$ $$\theta =n\pi ,n\pi +\pi /4,n\pi +\alpha $$ where $$tan\alpha =-2$$.

From given relation, if $$tan\theta =t$$

$$tan\theta =t=\dfrac { 2{ t }^{ 2 }-\dfrac { 1 }{ 3 } t }{ 1+\dfrac { 2 }{ 3 } { t }^{ 3 } } $$

$$\therefore $$ $$t=0$$ or $$1.\left( 1+\dfrac { 2 }{ 3 } { t }^{ 3 } \right) =2t-\dfrac { 1 }{ 3 } $$.

or $$2{ t }^{ 3 }-6t+4=0$$ or $${ t }^{ 3 }-3t+2=0$$

or $$(t+2)({t}^{2}-2t+1)=0$$

or $$(t+2){ \left( t-1 \right) }^{ 2 }=0$$

$$\therefore $$ $$t=tan\theta =-2,1\Rightarrow $$ (i),(ii).

(d) If $$t=tan\theta $$, then $$sin2\theta =\dfrac { 2t }{ 1+{ t }^{ 2 } } ,cos2\theta =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } $$

$$\therefore $$ $$\dfrac { 3sin2\theta }{ 5+4cos2\theta } =\dfrac { 3.2t }{ 5(1+{ t }^{ 2 })+4(1+{ t }^{ 2 }) } =\dfrac { 6t }{ 9+{ t }^{ 2 } } =\dfrac { 2.t/2 }{ 1+{ \left( t/3 \right) }^{ 2 } } =\dfrac { 2tan\alpha }{ 1+{ tan }^{ 2 }\alpha } =sin2\alpha $$

where $$\dfrac { t }{ 3 } =tan\alpha $$

$$\dfrac { 1 }{ 2 } { sin }^{ -1 }\left( \dfrac { 3sin2\theta }{ 5+4cos2\theta } \right) =\alpha ={ tan }^{ -1 }\left( \dfrac { t }{ 3 } \right) ={ tan }^{ -1 }\left( \dfrac { 1 }{ 3 } tan\theta \right) $$

$$\Rightarrow $$ $${ tan }^{ -1 }x={ tan }^{ -1 }\left( \dfrac { 1 }{ 3 } tan\theta \right) $$ $$\therefore $$ $$x=\dfrac { 1 }{ 3 } tan\theta $$

Simplify $${\cot ^{ - 1}}\dfrac{1}{{\sqrt {{x^2} - 1} }}$$ for $$x < - 1$$

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$$\cos ^{-1}x$$
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$$\sec^{-1}x$$
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$$\text{cosec}^{-1}x$$
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$$\tan ^{-1}x$$

Find value of $$\cos^{-1}\left(-\dfrac {1}{2}\right)$$.

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$$\dfrac {7\pi}{3}$$
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$$\dfrac {2\pi}{3}$$
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$$\dfrac {5\pi}{3}$$
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$$\dfrac {\pi}{3}$$

if f(x)=$${\tan ^{ - 1}}$$(x)than f(x)+f(y) is equal to:

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$${{\tan }^{-1}}x-{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$
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$${{\tan }^{-1}}x+{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$$
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$${{\tan }^{-1}}x+{{\tan }^{-1}}y$$=$${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$
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None of these

State True or False

$$\sin^{-1}2+\cos^{-1} 2=\dfrac{\pi}{2}$$.

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True
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False

For $$x\in(0,\pi/2) $$

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

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

$$\cos ^{-1}\left ( \cos \left ( \frac{5\pi}{4} \right ) \right )$$ is given by

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$$5\pi/4$$
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$$3\pi/4$$
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$$-\pi/4$$
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None of these

If $$\sin^{-1}\dfrac{1}{3} + \sin^{-1}\dfrac{2}{3} = \sin^{-1}x$$, then $$x$$ is equal to-

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$$0$$
0%
$$\dfrac{\sqrt{5} - 4\sqrt{2}}{9}$$
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$$\dfrac{\sqrt{5} + 4\sqrt{2}}{9}$$
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$$\dfrac{\pi}{2}$$

$$\tan^{-1} \dfrac{1}{2} + \tan^{-1} \dfrac{1}{3} $$ equals

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$$-\cfrac{\pi }{4}$$
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$$\cfrac{\pi }{6}$$
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$$\cfrac{\pi }{4}$$
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None of these

The value of $$\cos^{-1} (\cos 12) - \sin^{-1} (\sin 12)$$ is

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$$0$$
0%
$$\pi$$
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$$8 \pi - 24$$
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none of these

If $$f(x)=sin^{-1}(sinx)+cos^{-1}(sinx)$$ and $$\phi (x)=f(f(f(x))),$$ then $$\phi '(x)=$$

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1
0%
sin x
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0
0%
None of these

$$\tan^{-1}\dfrac {1}{5}+\tan^{-1}\dfrac {1}{7}+\tan^{-1}\dfrac {1}{3}+\tan^{-1}\dfrac {1}{8}$$ is equal to

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

If $$n - 1\sum\limits_{}^\infty {{{\cot }^{ - 1}}\left( {{{{n^2}} \over 8}} \right) = \pi .} $$ where $${a \over b}$$ is rational number in its lowest, then correct option is/are

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$$a - b = 3$$
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$$a + b = 11$$
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$$a + b = 10$$
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$$a - b = 4$$

$$\tan(\cot^{-1}x)$$ is equal to :

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$$\dfrac{\pi}{2}-x$$
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$$\cot(\tan^{-1}x)$$
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$$\tan x$$
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$$\dfrac1x$$

If $$\sin (\sin^{-1}\dfrac{1}{5}+\cos ^{-1}x)=1$$, then find the value of x.

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$$-1$$
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$$\dfrac { 1 }{ 3 } $$
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$$\dfrac { 1 }{ 5 } $$
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$$\dfrac { 1 }{ 2 } $$

The value of the expression $$2\sec^{-1} 2 +\sin^{-1} \dfrac{1}{2}$$ is

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$$\dfrac{\pi}{6}$$
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$$\dfrac{5\pi}{6}$$
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$$\dfrac{7\pi}{6}$$
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$$1$$

Number of solutions of the equation $$3{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \pi $$ is

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Zero
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$$2$$
0%
$$3$$
0%
$$1$$

If $$x,y,z \in [-1,1]$$ such that $$\cos^{-1}x +\cos^{-1}y +\cos^{-1}z=0$$, find $$x+y+z$$.

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

Find the value of :

$$\sec^2 (\tan^{-1} 2) +\csc^2 (\cot^{-1} 3)$$

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$$11$$
0%
$$15$$
0%
$$17$$
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$$21$$

Find the value of $$\sin^{-1}(2\cos^{2}x-1)+\cos^{-1}(1-2\sin^{2}x)$$.

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

Find the value of $$\cot (\tan^{-1} a +\cot^{-1} a)$$.

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

$$\sin^{-1}{0}$$ is equal to:

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

The value of $$x$$ for which $$\sin { \left( \cot ^{ -1 }{ \left( 1+x \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) } $$ is

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

Assertion (A) : The maximum value of $$f(x)=\sin^{-1}x+\cos^{-1}x+\tan^{-1}x$$ is $$\displaystyle \frac{3\pi}{4}$$
Reason (R) : $$\sin ^{-1} x>\cos^{-1}x$$ for all $$x$$ in $$R$$

0%
Both A and R are true and R is the correct explanation of A
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Both A and R are true and R is not correct explanation of A
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A is true but R is false
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A is false but R is true

The number of solutions of the equation

$$2(Sin^{-1}x)^{2}-5Sin^{-1}x+2=0$$ is

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0
0%
1
0%
2
0%
3

The number of triplets $$(x,y,z)$$ satisfying $$\sin^{-1}x+\sin^{-1}y+\cos^{-1}z=2\pi$$ is

0%
$$1$$
0%
$$0$$
0%
$$2$$
0%
$$\infty$$

Assertion ($$A$$) lf $$0<\displaystyle x<\frac{\pi}{2}$$ then $$\sin^{-1}(cosx)+\cos^{-1}(sinx)=\pi-2x$$
Reason (R) $$\displaystyle \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\forall x\in[0,1]$$

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Both A and R are true and R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but Ris false
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A is false but R is true

The smallest and the largest values of
$$\displaystyle \tan^{-1}\left (\dfrac{1-x}{1+x}\right)$$ , $$0\leq x\leq 1$$ are.

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

The value of $$x$$ where $$x>0$$ $$\displaystyle \tan(\sec^{-1}\frac{1}{x})=\sin(\tan^{-1}2)$$ is

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$$\sqrt{5}$$
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$$\displaystyle \frac{\sqrt{5}}{3}$$
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$$1$$
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$$\displaystyle \frac{2}{3}$$

The ascending order of $$A=\sin^{-1}(\log_{3}{2})$$ , $$B=\displaystyle \cos^{-1}\left(\log_{3}\left(\frac{1}{2}\right)\right)$$ , and $$C=\tan^{-1}\left(\log_{1/3} 2 \right)$$ is

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$$\text{C, B, A}$$
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$$\text{B, A, C}$$
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$$\text{C, A, B}$$
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$$ \text {B, C, A}$$

The equation 2$$\displaystyle \cos^{-1}x+\sin^{-1}x=\frac{11\pi}{6}$$ has

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No solution
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One solution
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Two solutions
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Three solutions

If $$x$$ takes negative permissible value, then $$\sin^{-1}x=$$

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$$\cos^{-1}\sqrt{1-x^{2}}$$
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$${-\cos^{-1}}\sqrt{1-x^{2}}$$
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$${\cos^{-1}}\sqrt{x^{2}-1}$$
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$${\pi-\cos^{-1}}\sqrt{1-x^{2}}$$

There is flag-staff at the top of $$10$$ metres high tower. lf the flag-staff makes an angle $$\tan ^{ -1 }{ \left( 1/8 \right) }$$ at a point $$24$$ metres away from the tower, then the height of the flag staff in metres is

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$$26/7$$
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$$27/8$$
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$$27/6$$
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$$26/3$$

If $$\tan(cos^{-1}x)=\sin (\sec^{-1} (\sqrt{5}) )$$, then $$x=$$

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

A tower stands at the top of a hill whose height is three times the height of the tower. The tower is found to subtend an angle of\$$\tan ^{ -1 }{ \left( { 1 }/{ 7 } \right) } $$ at a point $$2km$$ away on the horizontal throught the foot of the hill. Then the height of the tower is

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$$\displaystyle \frac{1}{2}km$$ or $$\displaystyle \frac{1}{3}km$$
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$$\displaystyle \frac{1}{3}km \space or \space \frac{2}{3}km$$
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$$\displaystyle \frac{2}{3}km \space or \space\frac{1}{2}km$$
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$$\displaystyle \frac{3}{4}km \space or \space \frac{1}{2}km$$

If $$\theta = sin^{-1}x+cos^{-1}x+tan^{-1}x,\ 0\leq x\leq 1$$, then the smallest interval in which $$\theta$$ lies is given by

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$$\displaystyle \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$$
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$$-\displaystyle \frac{\pi}{4}\leq\theta\leq 0$$
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$$0\displaystyle \leq\theta\leq\frac{\pi}{4}$$
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$$\displaystyle \frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}$$

The domain of $$\sin^{-1}[\log_{2}(x^{2}/2)]$$ is

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$$[2, 1]$$
0%
$$[1, 2]$$
0%
$$[-2,-1]\cup [1,2]$$
0%
$$[-2,0]$$

$$A$$ vertical pole subtends an angle $$\displaystyle \tan^{-1}(\frac{1}{2})$$ at a point $$P$$ on the ground. The angle subtended by the upper half of the pole at $$P$$ is

0%
$$\displaystyle \tan^{-1}(\frac{1}{4})$$
0%
$$\displaystyle \tan^{-1}(\frac{1}{8})$$
0%
$$\displaystyle \tan^{-1}(\frac{2}{3})$$
0%
$$\displaystyle \tan^{-1}(\frac{2}{9})$$

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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