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

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

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

$$\displaystyle \dfrac{\sqrt{5}}{3}$$

$$\displaystyle \dfrac{\sqrt{5}}{2}$$

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

$0\leq (sin^{-1}x-\dfrac {\pi}{4})^2\leq \dfrac {9\pi^2}{16}$ .

0%
Both statements I and II are true.
0%
Both statements I and II are true but I is not an explanation of II
0%
Statement I is true and statement II is false
0%
Statement I is false and statement II is true.

Explanation

Say

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

$f'\left( x \right) =0$ at

$x=\dfrac { \pi }{ 4 }$

$f''\left(x\right) \ge0$ at

$x=\dfrac{\pi}{4}$

$f\left( x \right) =\dfrac { { \pi }^{ 3 } }{ 32 }$ This is least value.

$\therefore f\left( x \right) \ge a{ \pi }^{ 3 }$ has a solution.

$\therefore \dfrac{1}{32}\ge$

$a$

Statement

$I$ is incorrect.

Now

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

$0\le { \left( \sin ^{ -1 }{ x } -\dfrac { \pi }{ 4 } \right) }^{ 2 }\le \dfrac { 9{ \pi }^{ 2 } }{ 16 }$

Min at

$x=\dfrac { \pi }{ 4 }$ and max at

$x=\dfrac { \pi }{ 2 }$ . So, Statement

$I$ is incorrect and

$II$ is correct.

$\cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } =\alpha$ , then

$4{ x }^{ 2 }-4xy\cos { \alpha } +{ y }^{ 2 }\quad$ is equal to

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

Explanation

Given,

$\cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } =\alpha$

$\Rightarrow \cos ^{ -1 }{ \left( \cfrac { xy }{ 2 } +\sqrt { \left( 1-{ x }^{ 2 } \right) \left( 1-\cfrac { { y }^{ 2 } }{ 4 } \right) } \right) =\alpha }$

$\Rightarrow \cos ^{ -1 }{ \left( \cfrac { xy+\sqrt { 4-{ y }^{ 2 }-4{ x }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 } } }{ 2 } \right) =\alpha } \quad$

$\Rightarrow 2\cos \alpha=xy+\sqrt {4-y^2-4x^2+x^2y^2}$

$\Rightarrow 4{ x }^{ 2 }+{ y }^{ 2 }-4xy\cos { \alpha } =4\sin ^{ 2 }{ \alpha }$

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

$x$ ?

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

Explanation

Given,

$\sin^{-1}\left (\dfrac {2a}{1 + a^{2}}\right ) - \cos^{1}\left (\dfrac {1 - b^{2}}{1 + b^{2}}\right ) = \tan^{-1}\left (\dfrac {2x}{1 - x^{2}}\right )$

$\Rightarrow 2\tan^{-1} a - 2\tan^{-1} b = 2\tan^{-1}x$

$\Rightarrow \tan^{-1}a - \tan^{-1}b = \tan^{-1}x$

$\Rightarrow \tan^{-1} \left (\dfrac {a - b}{1 + ab}\right ) = \tan^{-1}x$

$\Rightarrow x = \dfrac {a - b}{1 + ab}$

The value of

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

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{5\pi}{12}$
0%
$\dfrac{3\pi}{4}$
0%
$\dfrac{13\pi}{12}$

Explanation

The value of

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

$=\dfrac{\pi}{4}+\dfrac{2\pi}{3}-\dfrac{\pi}{6}$

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

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

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

Explanation

$\frac { x }{ 2 } \epsilon (-1,1)$ ,

$x\epsilon (-1,1)$
Intersection for

${ cos }^{ -1 }\left( \frac { x }{ 2 } \right)$ and

${ cos }^{ -1 }(x)$ to be valid
Thus x lies from (-1, 1)
for

$cos\left( { cos }^{ -1 }\left( \frac { x }{ 2 } \right) \right) =\frac { x }{ 2 }$

$\frac { x }{ 2 }$ should be from

$(2n\pi ,(2n+1))$
Similarly for

$cos({ cos }^{ -1 }x)=x$
x should be from

$(2n\pi ,(2n+1)\pi )$
Thus,

$x\epsilon (0,1)$

Solve:

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

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

Explanation

$\displaystyle { \tan }^{ -1 }\left( \frac { x }{ y } \right) -{ \tan }^{ -1 }\frac { x-y }{ x+y }$

$\displaystyle ={ \tan }^{ -1 }\left[ \frac { \frac { x }{ y } -\frac { x-y }{ x+y } }{ 1+\left( \frac { x }{ y } \right) \left( \frac { x-y }{ x+y } \right) } \right]$

$\displaystyle ={ \tan }^{ -1 }\left[ \frac { \frac { x\left( x+y \right) -y\left( x-y \right) }{ y\left( x+y \right) } }{ \frac { y\left( x+y \right) +x\left( x-y \right) }{ y\left( x+y \right) } } \right]$

$\displaystyle ={ \tan }^{ -1 }\left( \frac { { x }^{ 2 }+xy-xy+{ y }^{ 2 } }{ xy+{ y }^{ 2 }+{ x }^{ 2 }-xy } \right)$

$\displaystyle ={ \tan }^{ -1 }\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) ={ \tan }^{ -1 }1=\frac { \pi }{ 4 }$

Value of

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

0%
$\dfrac{\pi}{2}-1$
0%
$1-\dfrac{\pi}{4}$
0%
$2-\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{4}-1$

Explanation

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

$tan^{-1}\begin{Bmatrix}\dfrac{2\sin1\,cos\,1-[sin^21+\cos^21]}{\cos^21-sin^21}\end{Bmatrix}$

$tan^{-1}\begin{Bmatrix}\dfrac{-[\cos1-\sin1]^2}{(\cos 1-\sin 1)(cos1+sin1)}\end{Bmatrix}$

$tan^{-1}\begin{Bmatrix}\dfrac{\cos 1-\sin 1}{\cos 1+\sin 1}\end{Bmatrix}$ divide by

$cos1=tan^{-1}(tan(1-\dfrac{\pi}{4}))=1-\dfrac{\pi}{4}$

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

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

Explanation

$-\cfrac { \pi }{ 2 } \le { sin }^{ -1 }\theta \le \cfrac { \pi }{ 2 }$

And for

${ sin }^{ -1 }x+{ sin }^{ -1 }y+{ sin }^{ -1 }z=\cfrac { 3\pi }{ 2 }$

${ sin }^{ -1 }x={ sin }^{ -1 }y={ sin }^{ -1 }z=\cfrac { \pi }{ 2 }$

$\Rightarrow x=y=z=1$

Hence option 'A' is correct.

Sin-1

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

$\mathrm{x}=$

0%
$(^{\frac{63}{65})}$
0%
$\displaystyle \frac{56}{65}$
0%
$\displaystyle \frac{56}{63}$
0%
$\displaystyle \frac{16}{63}$

$\sec^{-1} x+ \sec^{-1}y + \sec^{-1}z = 3\pi$ , then

$xy + yz + zx =$ _______.

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

Explanation

Range of

$\sec^{-1}x\space :\space [0,\pi/2)\cup(\pi/2,\pi]$

$\sec^{-1}x+\sec^{-1}y+\sec^{-1}z=3\pi$ only when all of them are equal to

$\pi$

That is,

$\sec^{-1}x=\pi,\sec^{-1}y=\pi,\sec^{-1}z=\pi$

$\sec^{-1}x=\pi \rightarrow x=-1$

Similarly,

$y=-1,z=-1$

$\therefore \space xy+yz+zx=1+1+1=3$

Hence,

$(C)$

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

0%
Only one integer
0%
More than 2 integers
0%
Only two integers
0%
No integer

Explanation

We have the range of

$\tan^{-1}x$ is

$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .

Now,

$2\tan^{-1}x-\sin^{-1}x+\cot^{-1}x-\cos^{-1}x=\tan^{-1}x+(\tan^{-1}x+\cot^{-1}x)-(\sin^{-1}x+\cos^{-1}x)=\tan^{-1}x$ [Since

$\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$ and

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

$-\dfrac{\pi}{2}<\tan^{-1}x<\dfrac{\pi}{2}$

$\Rightarrow -1<\dfrac{2}{\pi}\tan^{-1}x<1$

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

$\Rightarrow -0.785<f(x)<0.785$

In this range

$f(x)$ has only

$1$ integer value and it is

$0$ .

The value of

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

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

Explanation

We need to find value of $\sin ^{ -1 }{ \left( \dfrac { 2\sqrt { 2 } }{ 3 } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) }$

Let $\sin ^{ -1 }{ \left( \dfrac { 2\sqrt { 2 } }{ 3 } \right) } =t$

$\Rightarrow \sin { t } =\dfrac { 2\sqrt { 2 } }{ 3 } \\ \sin ^{ 2 }{ t } +\cos ^{ 2 }{ t } =1\\ { \left( \dfrac { 2\sqrt { 2 } }{ 3 } \right) }^{ 2 }+\cos ^{ 2 }{ t } =1\\ \Rightarrow \cos ^{ 2 }{ t } =\dfrac { 1 }{ 9 } \\ \Rightarrow \cos { t } =\dfrac { 1 }{ 3 } \\ \Rightarrow t=\cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) }$

Substituting $t$

$\Rightarrow \cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) } +\sin ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) } \\ =\dfrac { \pi }{ 2 }$

So, option C is correct.

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

$xy<1$

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

$xy>1$ .

Evaluate:

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

where

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

0%
$\alpha$
0%
$2\alpha$
0%
$3\alpha$
0%
$4\alpha$

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

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

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

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

$\therefore$

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

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

$t(3+2{ t }^{ 3 })=t(6t-1)$

$\therefore$

$t=0$ or

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

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

$\therefore$

$t=0$ or

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

$t=0$ 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

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

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

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

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

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

$\therefore$

$t=tan\theta =-2,1\Rightarrow$ (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

$x < - 1$

0%
$\cos ^{-1}x$
0%
$\sec^{-1}x$
0%
$\text{cosec}^{-1}x$
0%
$\tan ^{-1}x$

Explanation

${\cot ^{ - 1}}\left( {\dfrac{1}{{\sqrt {{x^2} - 1} }}} \right)$

$let\ x = {\sec ^2}\theta$

${\cot ^{ - 1}}\left( {\dfrac{1}{{\sqrt {{{\sec }^2}\theta-1 }}}} \right)$

$1 + {\tan ^2}\theta = {\sec ^2}\theta$

$\therefore {\sec ^2}\theta - 1 = {\tan ^2}\theta$

${\cot ^{ - 1}}\left( {\dfrac{1}{{\sqrt {{{\tan }^2}\theta } }}} \right)$

${\cot ^{ - 1}}\left( {\dfrac{1}{{\tan \theta }}} \right)$

${\cot ^{ - 1}}\left( {\cot \theta } \right)=\theta$

$={\sec ^{ - 1}}x.$

Find value of

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

0%
$\dfrac {7\pi}{3}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {5\pi}{3}$
0%
$\dfrac {\pi}{3}$

Explanation

$\begin{array}{l} { \cos ^{ -1 } }\left( { \dfrac { { -1 } }{ 2 } } \right) \\ =\pi -{ \cos ^{ -1 } }\left( { \dfrac { 1 }{ 2 } } \right) \\ =\pi -{ \cos ^{ -1 } }\left( { \cos \dfrac { \pi }{ 3 } } \right) \\ =\pi -\dfrac { \pi }{ 3 } \\ =\dfrac { { 2\pi } }{ 3 } \end{array}$

if f(x)=

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

0%
${{\tan }^{-1}}x-{{\tan }^{-1}}y$ = ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$
0%
${{\tan }^{-1}}x+{{\tan }^{-1}}y$ = ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$
0%
${{\tan }^{-1}}x+{{\tan }^{-1}}y$ = ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$
0%
None of these

Explanation

Given that , $f\left( x \right)={{\tan }^{-1}}x$ ….(1)

Put x=y in equation (1) we get

$f\left( y \right)={{\tan }^{-1}}y$ ….(2)

Add equation (1) and (2)

$f\left( x \right)+f\left( y \right)={{\tan }^{-1}}x+{{\tan }^{-1}}y$

Using inverse trigonometry properties, we get

${{\tan }^{-1}}x+{{\tan }^{-1}}y$ $={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$

This is the answer

State True or False

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

0%
True
0%
False

Explanation

We have

$\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$ only when

$x\in [-1,1]$ .

Since

$2\not\in[-1,1]$ , so

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

For

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

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

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

Explanation

We have,

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

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

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

Hence, this is the answer.

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

0%
$5\pi/4$
0%
$3\pi/4$
0%
$-\pi/4$
0%
None of these

Explanation

$\cos^{-1}(\cos(5\pi/4))$

$=\cos^{-1}(\cos(\pi+\pi/4))$

$\cos^{-1}(\cfrac{-1}{\sqrt{2}})$

$=\cfrac{5\pi}{4}$

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

$x$ is equal to-

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

Explanation

Given

$sin^{-1}\dfrac{1}{3}+sin^{-1}\dfrac{2}{3} = sin^{-1}x$

To find x

Sol :-

$sin^{-1}\dfrac{1}{3}+sin^{-1}\dfrac{2}{3}$

$= sin^{-1}\left ( \dfrac{1}{3}\sqrt{1-\dfrac{4}{9}}+\dfrac{2}{3}\sqrt{1-\dfrac{1}{9}} \right )$

$\left [ sin^{-1}x+sin^{-1}y = sin^{-1}(x+\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}) \right ]$

$= sin^{-1}\left ( \dfrac{1}{3}\dfrac{\sqrt{5}}{3}+\dfrac{2}{3}\dfrac{\sqrt{8}}{3} \right )$

$= sin^{-1}\left ( \dfrac{\sqrt{5}+4\sqrt{2}}{9} \right )$

$\because sin^{-1}\dfrac{1}{3}+sin^{-1}\dfrac{2}{3} = sin^{-1}x$ (Given)

$\therefore x = \dfrac{\sqrt{5}+4\sqrt{2}}{9}$

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

0%
$-\cfrac{\pi }{4}$
0%
$\cfrac{\pi }{6}$
0%
$\cfrac{\pi }{4}$
0%
None of these

Explanation

Let

$\tan^{-1}\cfrac{1}{2}=x$

$\Rightarrow \tan x=\cfrac{1}{2}$

Let

$\tan^{-1}\cfrac{1}{3}=y$

$\Rightarrow \tan y=\cfrac{1}{3}$

$\tan (x+y)=\cfrac{\tan x + \tan y}{1-\tan x.\tan y}$

$=\cfrac{\cfrac{1}{2}+\cfrac{1}{3}}{1-\cfrac{1}{2}\times \cfrac{1}{3}}$

$=1$

$x+y=\tan^{-1}1=\cfrac{\pi }{4}$

$\tan^{-1}\cfrac{1}{2}+\tan^{-1}\cfrac{1}{3}=\cfrac{\pi }{4}$

The value of

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

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

Explanation

$12\,rad$ lies in

$4th$ quadrant

$\dfrac{7\pi}{2}<12<4\pi$
Let

$\theta$ be an acute angle such that

$12+\theta=4\pi$

$\therefore 12=4\pi-\theta$ or

$\theta=4\pi-12$

$\cos^{-1}(\cos12)-\sin^{-1}(\sin12)$

$=\cos^{-1}(\cos(4\pi-\theta))-\sin^{-1}(\sin(4\pi-\theta))$

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

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

$=\theta-(-\theta)$

$=2\theta$

$=2(4\pi-12)$

$=8\pi-24$

$\boxed{\therefore\,\cos^{-1}(\cos12)-\sin^{-1}(\sin12)=8\pi-24}$

$f(x)=sin^{-1}(sinx)+cos^{-1}(sinx)$ and

$\phi (x)=f(f(f(x))),$ then

$\phi '(x)=$

0%
1
0%
sin x
0%
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

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

Explanation

$\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{7}+\tan^{-1}\dfrac{1}{3}+\tan^{-1}\left(\dfrac{1}{8}\right)$

$\tan^{-1}\left(\dfrac{1}{5}\right)+\tan^{-1}\left(\dfrac{1}{3}\right)+\tan^{-1}\left(\dfrac{1}{7}\right)+\tan^{-1}\left(\dfrac{1}{8}\right)$

$\Rightarrow \tan^{-1}\left(\dfrac{8}{14}\right)+\tan^{-1}\left(\dfrac{15}{55}\right)$

$\Rightarrow \tan^{-1}\left(\dfrac{4}{7}\right)+\tan^{-1}\left(\dfrac{3}{11}\right)$

$\Rightarrow \tan^{-1}\left(\dfrac{44+21}{77-12}\right)=\tan^{-1}\left(\dfrac{65}{65}\right)$

$\Rightarrow \tan^{-1}(1)=\dfrac{\pi}{4}$ .

1190628_1155201_ans_d56c1f57a37c46059a7b54312d670191.jpg

$n - 1\sum\limits_{}^\infty {{{\cot }^{ - 1}}\left( {{{{n^2}} \over 8}} \right) = \pi .}$ where

${a \over b}$ is rational number in its lowest, then correct option is/are

0%
$a - b = 3$
0%
$a + b = 11$
0%
$a + b = 10$
0%
$a - b = 4$

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

0%
$\dfrac{\pi}{2}-x$
0%
$\cot(\tan^{-1}x)$
0%
$\tan x$
0%
$\dfrac1x$

Explanation

$\tan(\cot^{-1}x)$

$=\tan \left(\tan^{-1}\left({\dfrac1x}\right)\right)$

$=\dfrac1x$

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

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

Explanation

$\sin(\sin^{-1}\dfrac{1}{5}+\cos^{-1}x)=1$

$\sin^{-1}\dfrac{1}{5}+\cos^{-1}x =\sin^{-1} 1$

$\sin^{-1}\dfrac{1}{5}+\cos ^{-1}x =\dfrac{\pi}{2}$

$\cos ^{-1} x =\dfrac{\pi}{2}-\sin^{-1}\dfrac{1}{5}$

$\cos ^{-1} x =\cos ^{-1}\dfrac{1}{5}$

$x=\dfrac{1}{5}$

The value of the expression

$2\sec^{-1} 2 +\sin^{-1} \dfrac{1}{2}$ is

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{5\pi}{6}$
0%
$\dfrac{7\pi}{6}$
0%
$1$

Explanation

$2\sec^{-1} 2 +\sin^{-1} \dfrac{1}{2}=\dfrac{2\pi}{3}+\dfrac{\pi}{6}=\dfrac{5\pi}{6}$

Number of solutions of the equation

$3{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \pi$ is

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

$x,y,z \in [-1,1]$ such that

$\cos^{-1}x +\cos^{-1}y +\cos^{-1}z=0$ , find

$x+y+z$ .

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

Explanation

We have,

$x,y,z \in [-1,1]$

$-1 \leq x \leq 1, -1\leq y \leq 1, -1\leq z \leq 1$

$0 \leq \cos ^{-1}x \leq \pi, 0 \leq \cos^{-1} y \leq \pi, 0 \leq \cos^{-1} z \leq \pi$

Therefore,

$\cos^{-1}x +\cos^{-1} y +\cos^{-1} z =0$

$\implies \cos^{-1} x=0, \cos^{-1} y=0, \cos^{-1} z=0$

$x=y=z=1$

Hence,

$x+y+z=3$

Find the value of :

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

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

Explanation

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

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

$=1+[\tan (\tan^{-1} 2)]^2 +1+[\cot (\cot^{-1} 3)]^2=1+2^2+1+3^2=15$

Find the value of

$\sin^{-1}(2\cos^{2}x-1)+\cos^{-1}(1-2\sin^{2}x)$ .

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

Explanation

As we know that

$\cos 2 A=2\cos^2 A-1=1-2\sin^2 A$

And also

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

$\text{sin}^{-1}(2\cos^2 x-1)+\text{cos}^{-1}(1-2\sin^2 x)=\text{sin}^{-1} (\cos 2 x)+\text{cos}^{-1} (\cos 2 x)=\dfrac{\pi}{2}$

Find the value of

$\cot (\tan^{-1} a +\cot^{-1} a)$ .

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

Explanation

We know,

$\tan^{-1} a +\cot^{-1} a=\dfrac{\pi}{2}$

Therefore,

$\cot (\tan^{-1} a +\cot^{-1} a )=\cot \dfrac{\pi}{2}=0$

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

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

Explanation

As we know that,

$\sin{0} = 0$

$\Rightarrow 0 = \sin^{-1}{\left( 0 \right)}$

Hence the value of

$\sin^{-1}{\left( 0 \right)}$ is

$0$ .

The value of

$x$ for which

$\sin { \left( \cot ^{ -1 }{ \left( 1+x \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) }$ is

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

Explanation

We have,

$\displaystyle \sin { \left( \cot ^{ -1 }{ \left( x+1 \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) }$

$\displaystyle \Rightarrow \cos { \left( \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } \right) } =\cos { \left( \tan ^{ -1 }{ x } \right) }$

$\displaystyle \Rightarrow \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } =2n\pi \pm \tan ^{ -1 }{ x }$

Put

$\displaystyle n=0\Rightarrow \frac { \pi }{ 2 } -\cot ^{ -1 }{ \left( x+1 \right) } =\pm \tan ^{ -1 }{ x } =\tan ^{ -1 }{ \left( \pm x \right) }$

$\displaystyle \Rightarrow \frac { \pi }{ 2 } =\tan ^{ -1 }{ \left( \pm x \right) } +\cot ^{ -1 }{ \left( x+1 \right) }$

$\displaystyle \Rightarrow x+1=\pm x\Rightarrow 2x+1=0$

$\displaystyle \therefore x=-\frac { 1 }{ 2 }$

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

Explanation

$f(x)=sin^{-1}(x)+cos^{-1}(x)+tan^{-1}(x)$
Now

$sin^{-1}(x)+cos^{-1}(x)=\dfrac{\pi}{2}$
Hence

$f(x)=tan^{-1}(x)+\dfrac{\pi}{2}$
Substituting x=1, we get the maximum as

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

$=\dfrac{3\pi}{4}$
Reason.
The intersection point of sin(x) and cos(x) is

$x=\dfrac{\pi}{4}$ between

$[0,\dfrac{\pi}{2}]$ .
Now cos(x) is a decreasing function in the above interval while sin(x) is an increasing function in the above interval.
Also.

$cos(x)>sin(x)$ between

$[0,\dfrac{\pi}{4}]$ while

$sin(x)>cos(x)$ between

$[\dfrac{\pi}{4},\dfrac{\pi}{2}]$
Hence

$sin^{-1}(x)>cos^{-1}(x)$ between

$[\dfrac{1}{\sqrt{2}},1]$ .
Hence reason is false.

The number of solutions of the equation

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

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

Explanation

Let

$sin ^{-1}x=t , t\epsilon [-\frac{\pi }{2} ,\frac{\pi }{2}]$

$\therefore 2(sin ^{-1}x)^{2}-5 sin ^{2}x+2=0$

$2t^{2}-5t+2=0$

$\displaystyle t=\frac{5\pm \sqrt{25-16}}{4}$

$\displaystyle =\frac{5\pm 3}{4}$

$t=2, \frac{1}{2}$

$sin ^{-1}x=2, \frac{1}{2}$

$sin ^{-1}x=2 or sin ^{-1}x=\frac{1}{2}$

$x=sin (\frac{1}{2}) and x=sin2$ satisfy the equations.

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%
$\infty$

Explanation

$\sin ^{-1}x+\sin ^{-1}y+\cos ^{-1}z=2\pi$

We have,

$\dfrac{-\pi}{2}\leq \sin^{-1} x \leq \dfrac {\pi}{2}$

$\dfrac{-\pi}{2}\leq \sin^{-1} y \leq \dfrac {\pi}{2}$

and

$0 \leq\cos^{-1} z \leq \pi$

thus for

$RHS$ to become

$2\pi$ we need to take the maximum values only where,

$\sin^{-1}x=\sin^{-1}y=\dfrac{\pi}{2}$

$\cos^{-1}z=\pi$

Hence,

$x = y = 1$

$z = -1$

$\therefore (x,y,z)=(1,1,-1)$

There is only one triplet satisfying the given equation.
Hence, option A is correct.

Assertion (

$A$ ) lf

$0<\displaystyle x<\frac{\pi}{2}$ then

$\sin^{-1}(cosx)+\cos^{-1}(sinx)=\pi-2x$
Reason (R)

$\displaystyle \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\forall x\in[0,1]$

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

Explanation

Given

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

$\sin ^ {-1} \left(\sin \left(\dfrac\pi 2- x\right)\right) +\cos ^{-1} \left(\cos \left(\dfrac\pi 2- x\right)\right)$

$\bigg( \sin \theta =\cos \left(\dfrac\pi 2- \theta \right) \bigg)$

$\dfrac \pi 2-x +\dfrac \pi 2-x$

$\pi -2x$

We know

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

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

The smallest and the largest values of

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

$0\leq x\leq 1$ are.

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

Explanation

$\displaystyle tan ^{-1}(\frac{1-x}{1+x})$

$0 \leq x\leq 1$

$1\leq x+1 \leq 2$

$1\geq \frac{1}{1+x} \geq \frac{1}{2}$

$\displaystyle 0 \leq \frac{1-x}{1+x}\leq 1$

$\therefore x\epsilon [-1,1] \space by\space x\epsilon [0,1]$

So,

$x\epsilon [0,1]$

$\displaystyle tan ^{-1}(\frac{1-x}{1+x}) \space \epsilon [0,\frac{\pi }{4}]$

The value of

$x$ where

$x>0$

$\displaystyle \tan(\sec^{-1}\frac{1}{x})=\sin(\tan^{-1}2)$ is

0%
$\sqrt{5}$
0%
$\displaystyle \frac{\sqrt{5}}{3}$
0%
$1$
0%
$\displaystyle \frac{2}{3}$

Explanation

$tan(sec^{-1}{\dfrac{1}{x}})=sin(tan^{-1}(2))$

$tan(tan^{-1}(\dfrac{\sqrt{1-x^2}}{x}))=sin(sin^{-1}(\dfrac{2}{\sqrt{5}}))$

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

$\dfrac{1-x^2}{x^2}=\dfrac{4}{5}$

$\dfrac{1}{x^2}-1=\dfrac{4}{5}$

$\dfrac{1}{x^2}=\dfrac{9}{5}$

$x^2=\dfrac{5}{9}$

$x=\pm\dfrac{\sqrt{5}}{3}$
Since

$x>0$

$x=\dfrac{\sqrt{5}}{3}$

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

0%
$\text{C, B, A}$
0%
$\text{B, A, C}$
0%
$\text{C, A, B}$
0%
$\text {B, C, A}$

Explanation

We have,

$A = \displaystyle \frac{1}{2} < \ log_{3}{2} < 1, \quad \because \sqrt 3 < 2 <3$

$\Rightarrow \displaystyle \frac{\pi}{6} < \sin^{-1} \left( \log_3 2\right) < \frac{\pi}{2}$

$\displaystyle \because \sin^{-1}{\frac{1}{2}} = \frac{\pi}{6}$ ,

$\displaystyle \sin^{-1} {1} = \frac{\pi}{2}$

$\displaystyle B=\cos^{-1}(\log_{3}{1/2})=\cos^{-1}\left(-\log_{3}{2}\right)$

$=\pi -cos ^{-1}(\log _{3}{2}) \quad [\cos^{-1}1/2 =60^{o} , \cos^{-1}(1)=0^{\circ}]$

$C=\tan^{-1}(\log _{1/3}{2})=\tan^{-1}(-\log_{3}{2})=- \tan^{-1}(\log_{3}{2}) \quad [\tan^{-1}(1/2) > 0^{\circ} , \tan^{-1}(1)=45^{\circ}]$

$C \text{ is } negative , 30^{o} < A < 90^{0} , \dfrac{2\pi }{3} < B < \pi$
Arrange By asscending order

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

The equation 2

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

0%
No solution
0%
One solution
0%
Two solutions
0%
Three solutions

Explanation

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

$\displaystyle \therefore cos ^{-1}x=\frac{11\pi }{6}-\frac{\pi }{2}$ Since

$[sin ^{-1}x+cos ^{-1}x=\frac{\pi }{2}]$

$\displaystyle cos ^{-1}x=\frac{8\pi }{6}=\frac{4\pi }{3}$
But range of

$cos^{-1}x\epsilon [0,\pi ]$
So, this equation has no solution.

$x$ takes negative permissible value, then

$\sin^{-1}x=$

0%
$\cos^{-1}\sqrt{1-x^{2}}$
0%
${-\cos^{-1}}\sqrt{1-x^{2}}$
0%
${\cos^{-1}}\sqrt{x^{2}-1}$
0%
${\pi-\cos^{-1}}\sqrt{1-x^{2}}$

Explanation

$-1\leq x <0$ , then

$\dfrac{-\pi}{2} \leq sin^{-1}(x) < 0$
Let,

$sin \theta = x$
Hence,

$\dfrac{-\pi}{2} \leq \theta < 0$
Hence,

$cos \theta = \sqrt{1-x^2}$
Since, the range of

$cos^{-1}(x)$ is

$[0,\pi]$ ,

$\theta = - cos^{-1} ( \sqrt{1-x^2} )$

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

0%
$26/7$
0%
$27/8$
0%
$27/6$
0%
$26/3$

Explanation

$tan\theta =\frac{1}{8}, tan\alpha =\frac{10}{24}$

$=\frac{5}{12}$

$tan(\alpha +\theta )= \frac{h+10}{24}$

$\frac{tan\alpha +tan\theta} {1-tan\alpha tan\theta }=\frac{h+10}{24}$

$\frac{1}{8}+\frac{5}{12}1-\frac{5}{12} \times\frac{1}{8}=\frac{h+10}{24}$

$\frac{12+40}{(69-5)}=\frac{h+10}{24}$

$\frac{25\times24}{91}=4+10$

$h=\frac{338}{91}= \frac{26}{7}$

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

$x=$

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

Explanation

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

$\therefore Let \space \alpha =cos ^{-1}x$

$cos \alpha =x$

$\displaystyle \therefore tan \alpha =\frac{\sqrt{1-x^{2}}}{x}$

$\displaystyle \therefore tan (cos ^{-1}x)=\frac{\sqrt{1-x^{2}}}{x}=\dfrac{2}{\sqrt{5}}$

$\Rightarrow 5(1-x^{2})=4x^{2}$

$5=9x^{2}$

$\displaystyle x=\pm\frac{\sqrt{5}}{3}$

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

0%
$\displaystyle \frac{1}{2}km$ or $\displaystyle \frac{1}{3}km$
0%
$\displaystyle \frac{1}{3}km \space or \space \frac{2}{3}km$
0%
$\displaystyle \frac{2}{3}km \space or \space\frac{1}{2}km$
0%
$\displaystyle \frac{3}{4}km \space or \space \frac{1}{2}km$

Explanation

$tan\theta =\frac{1}{7}$

$tan\theta =\frac{3h}{2}$

$(tan\theta+\alpha)=\frac{4h}{2}$

$\frac{4tan\alpha}{3}=\frac{tan\theta+tan\alpha}{1-tan\alpha tan\theta}$

$\frac{4tan\alpha}{3}=\frac{1+7tan\alpha}{7-tan\alpha}$

$28tan\alpha-4tan^{2}\theta=3+21tan\theta$

$4tan^{2}\theta-7tan\theta =3+21tan\theta$

$tan\theta=1,\frac{3}{4}$

$h=\frac{2}{3},1/2$

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

$\theta$ lies is given by

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

Explanation

$\theta = sin^{-1}x + cos^{-1} x +tan^{-1} x$
Now,

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

$\Rightarrow \theta = \dfrac{\pi}{2} + tan^{-1}{x}$
For ,

$0 \leq x \leq 1$

$0 \leq tan^{-1} x \leq \dfrac{\pi}{4}$

$\Rightarrow \dfrac{\pi}{2} \leq \theta \leq \dfrac{3\pi}{4}$
Hence, option D is correct.

The domain of

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

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

Explanation

$\sin ^{-1}(x) domain : x\in [-1,1]$

$\therefore$ domain of

$\displaystyle \sin ^{-1}[\log _{2}\dfrac{x^{2}}{2}]$

$\therefore \log_{2}\dfrac{x^{2}}{2} \in [-1,1]$

$\displaystyle -1 \leq \log_{2}\frac{x^{2}}{2} \leq 1$ and

$\displaystyle \frac{1}{2} \leq \frac{x^{2}}{2} \leq 2$

$1\leq x^{2} \leq 4$

For

$x^{2} \geq 1$

$x\in [-\infty ,-1] \cup [1, \infty ]$ ...........(i)

For

$x^{2}\leq 4$

$x\in [-2,2]$ .................(ii)

By taking intersection of (i) & (ii)

$\therefore x\epsilon [-2,-1] \cup [1,2]$

$A$ 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%
$\displaystyle \tan^{-1}(\frac{2}{9})$

Explanation

as shown in the figure

$tan(tan^{-1}(\dfrac{1}{2}))=\dfrac{1}{2}=\dfrac{2h}{b}$ .......................(1)

$tan(tan^{-1}(\dfrac{1}{2})-\phi)=\dfrac{h}{b}$ .......................(2)

(1) divided by (2) we get

$2= \dfrac{\dfrac{1}{2}}{\dfrac{1-\dfrac{tan\phi}{2}}{1+\dfrac{tan\phi}{2}}}$

solving the above equation we get

$tan\phi= \dfrac{2}{9}$

$\phi= tan^{-1}\dfrac{2}{9}$

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

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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