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

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

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

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

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

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%
$\dfrac{\pi}{2}$

Explanation

Solution:-

$sin^{-1}\left \{ cot \left \{ sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+cos^{-1}\dfrac{\sqrt{12}}{4}+sec^{-1}\sqrt{2} \right \} \right \}$

We know that

$sin \dfrac{\pi}{12} = \sqrt{\dfrac{2-\sqrt{3}}{4}} cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{12}}{4}$

$sec \dfrac{\pi}{4} = \sqrt{2}$

$sin^{-1} (cot(\dfrac{\pi}{12}+\dfrac{\pi}{6} + \dfrac{\pi}{4}))$

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

$sin^{-1}0$

$0$

C is correct.

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

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

Explanation

$\tan^-1 \sqrt {3}=\dfrac{\pi}{3}$

and

$\cot^-1(-\sqrt{3})=\left(\pi-\dfrac{\pi}{6}\right)=\dfrac{5\pi}{6}$

$\therefore$

$\dfrac{\pi}{3}-\dfrac{5\pi}{6}=\dfrac{-3\pi}{6}=\dfrac{-\pi}{2}$

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

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

Explanation

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

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

$\displaystyle = \cos^{-1}\left \{ \cos\left ( \frac{\pi }{4}+\frac{9\pi }{10} \right ) \right \}$

$\displaystyle =\cos^{-1}\left \{ \cos\left ( \frac{23\pi }{20} \right ) \right \}$

$=\dfrac{23\pi }{20}$

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%
$[-1, 1]$

Explanation

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

let

$x = \cos \theta ,\,\theta = {\cos ^{ - 1}}x$

$\begin{array}{l} \frac { \pi }{ 2 } ={ \cot ^{ -1 } }\frac { { \sqrt { 1-{ { \cos }^{ 2 } }\theta } } }{ { \cos \theta } } -{ \sin ^{ -1 } }\sqrt { 1-{ { \cos }^{ 2 } }\theta } \\ \frac { \pi }{ 2 } ={ \cot ^{ -1 } }\left( { \frac { { \sin \theta } }{ { \cos \theta } } } \right) -{ \sin ^{ -1 } }\left( { \sin \theta } \right) \\ \frac { \pi }{ 2 } ={ \cot ^{ -1 } }\left[ { \cot \left( { \frac { \pi }{ 2 } -\theta } \right) } \right] -\theta \\ \frac { \pi }{ 2 } =\frac { \pi }{ 2 } -\theta -\theta \\ 0=-2\theta \\ \theta =0 \\ { \cos ^{ -1 } }x=0 \\ x=\cos { 0^{ 0 } } \\ x=1 \\ \left[ { -1,1 } \right] -\{ 0\} \end{array}$

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%
$\sqrt{3}$

Explanation

$\tan \left[\sin^{-1}\left(\dfrac{x}{\sqrt2} + \sqrt{\dfrac{1-x^2}{2}}\right)-\sin x\right]$

Put

$x = \sin \theta$

$\tan \left[\sin^{-1}\left(\dfrac{\sin \theta + \cos \theta}{\sqrt{2}}\right)+(-\theta)\right]$

$= \tan \left[\sin^{-1}\left(\sin \theta \cos\left(\dfrac{\pi}{4} \right)+\cos \theta\sin\left(\dfrac{\pi}{4} \right)\right)-\theta\right]$

$= \tan \left[\sin^{-1}\sin\left(\dfrac{\pi}{4} + \theta\right)-\theta\right]$

$= \tan \left(\dfrac{\pi}{4}\right)$

$= 1$

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%
$\tan ^{ -1 }{ \left( \dfrac { 44 }{ 125 } \right) }$

Explanation

$2\cot^{-1}7+\cos^{-1}\left(\dfrac{3}{5}\right)$

Let

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

$\cos\theta =\dfrac{3}{5}$

$\cot\theta =\dfrac{3}{4}$

$2\cot^{-1}(7)+\cot^{-1}\left(\dfrac{3}{4}\right)$

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

$\cot^{-1}\left(\dfrac{7^2-1}{2\times 7}\right)+\cot^{-1}\left(\dfrac{3}{4}\right)$

$\cot^{-1}\left(\dfrac{24}{7}\right)+\cot^{-1}\left(\dfrac{3}{4}\right)$

$\cot^{-1}\left(\dfrac{\dfrac{24}{7}\times \dfrac{3}{4}-1}{\dfrac{24}{7}+\dfrac{3}{4}}\right)$

$\left[\because \cot^{-1}(a)+\cot^{-1}(b)=\cot^{-1} \left(\dfrac{ab-1}{a+b}\right)\right]$

$\cot^{-1}\left(\dfrac{44/28}{117/28}\right)=\cot^{-1}\left(\dfrac{44}{117}\right)$ .

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%
$\dfrac {\pi}{2}$

Explanation

$\tan^{-1}\left(2\sin\left(\sec^{-1}(2)\right)\right)$

Let

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

$\tan^{-1}\left(2\sin x\right)$ ----- ( 1 )

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

$\Rightarrow$

$\sec x=2$

$\Rightarrow$

$\dfrac{1}{\cos x}=2$

$\Rightarrow$

$\dfrac{1}{2}=cos x$

$\Rightarrow$

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

$\Rightarrow$

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

Substituting value of

$x$ in ( 1 ) we get,

$\Rightarrow$

$\tan^{-1}\left(2\sin \dfrac{\pi}{3}\right)$

$\Rightarrow$

$\tan^{-1}\left(2\times \dfrac{\sqrt{3}}{2}\right)$

$\Rightarrow$

$\tan^{-1}(\sqrt{3})$

$\Rightarrow$

$\dfrac{\pi}{3}$

$\therefore$

$\tan^{-1}\left(2\sin\left(\sec^{-1}(2)\right)\right)=\dfrac{\pi}{3}$

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

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

Explanation

$cot^{-1}(\dfrac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}})$

$sin 2x = 2sinx cosx.$

also,

$sinx = 2sin \dfrac{x}{2} cos\dfrac{x}{2} ...(1)$

also,

$1+sinx = 2sin\dfrac{x}{2} cos \dfrac{x}{2}+1$

since

$sin^2x+cos^2x = 1$

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

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

$1+sinx = (sin\dfrac{x}{2}+cos \dfrac{x}{2})^2$

$\sqrt{1+sinx} = sin\dfrac{x}{2}+cos\dfrac{x}{2} ...(2)$

Similarly

multiply by -1 in

$eq^n$ (1),

$-sinx = 2sin\dfrac{x}{2} cos\dfrac{x}{2}$

Now adding 1 to the above

$eq^n$

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

$\therefore$ we get

$\sqrt{1-sinx} = (cos\dfrac{x}{2}-sin\dfrac{x}{2})...(3)$

$cos^{-1}(\dfrac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}})$

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

$= cos^{-1}(\frac{2cos\dfrac{x}{2}}{-2sin\dfrac{x}{2}}) = cot^{-1}(cot\dfrac{x}{2})$

$= -cot^{-1}(cot\dfrac{x}{2}) (\therefore cotx$ is an add function)

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

1159086_1139085_ans_b3a393698543444e957f6ca653cfb2f6.jpg

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%
$5000$

Explanation

Given

Income of

$A$ is

$8000$

Income of

$B$ is

$9000$

Expense Ratio is

$5:3$

Let the common factor be

$x$

So Expenses of

$A$ are

$5x$

Expenses of

$B$ are

$3x$

Savings Ratio is

$1:2$

Let the common factor be

$y$

So Savings of

$A$ are

$y$

Savings of

$B$ are

$2y$

According to Question

$5x+y=8000\cdots(1)$

$3x+2y=9000\cdots(2)$

$2(1)-(2)$

$\implies 10x+2y-3x-2y=16000-9000$

$\implies 7x=7000$

$\implies x=1000$

Expenses of

$A$ is

$5x=5\times 1000=5000$

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

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.

$\sin^{-1}\left(x-\dfrac {x^{2}}{2}+\dfrac {x^{3}}{4}-...\infty \right)+\cos^{-1}\left(x^{2}-\dfrac {x^{4}}{2}+\dfrac {x^{6}}{4}-...\infty \right)=\dfrac {\pi}{2}$ for

$0 < |x| < \sqrt {2}$ ,then

$x$ equal

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

Explanation

As we know that

Sum of an infinite

$GP$ is

$\dfrac{a}{1-r}$ for

$|r|<1$

$x-\dfrac{x^2}{2}+\dfrac{x^3}{4}-\cdots\infty=\dfrac{x}{1+\dfrac{x}{2}}$

And also

$x^2-\dfrac{x^2}{2}+\dfrac{x^6}{4}-\cdots\infty=\dfrac{x^2}{1+\dfrac{x^2}{2}}$

$\text{sin}^{-1}\bigg(x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots\infty\bigg)+\text{cos}^{-1}\bigg(x^2-\frac{x^4}{4}+\frac{x^6}{4}-\cdots\infty\bigg)=\dfrac{\pi}{2}$

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

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

$\implies \dfrac{1}{x+2}=\dfrac{x}{x^2+2}$

$\implies x^2+2 x=x^2+2$

$\implies 2 x=2$

$\implies x=1$

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

0%
True
0%
False

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

Explanation

We know that

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

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

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

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

$={\tan}^{-1}{\left(\dfrac{7+3}{21-1}\right)}+{\tan}^{-1}{\left(\dfrac{13}{40-1}\right)}\\$

$={\tan}^{-1}{\left(\dfrac{10}{20}\right)}+{\tan}^{-1}{\left(\dfrac{13}{39}\right)}\\$

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

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

$={\tan}^{-1}{\left(\dfrac{5}{5}\right)}={\tan}^{-1}{1}=\dfrac{\pi}{4}\neq \dfrac{\pi}{2}\\$

Hence the given statement is 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

Explanation

Given

$\cos \left(\tan ^{-1} \left(\tan \dfrac {15\pi}4 \right)\right)$

$\implies \cos \left(\tan ^{-1} \left(\tan \left(4\pi -\dfrac {\pi}4 \right)\right)\right)$

$\implies \cos \left(\tan ^{-1} \left(\tan \left(-\dfrac {\pi}4 \right)\right)\right)$

$\bigg(\tan (-x)=-\tan x\bigg)$

$\implies \cos \left(-\dfrac {\pi}4 \right)$

$\implies \dfrac 1{\sqrt 2}$

If the number

$93215x2$ is completely divisible by

$11$ , then

$x$ is equal to

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

Explanation

Given number is

$93215x2$

The number can be represented as

$9321502+10x$

Dividing

$9321502$ leaves

$3$ as remainder

So the expression

$10x+3$ should be divisible by

$11$ to make the whole number to be divisible by

$11$

$\implies 10x+3=11m(say)$

$\implies x=\dfrac{11m-3}{10}$

The expression gives an integer value at

$m=3$

$\implies x=\dfrac{33-3}{10}$

$\implies x=3$

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%
$\dfrac {\pi}{6}$

Explanation

Let

$y=\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$

$\cos y=\cos\left(\dfrac{7\pi}{6}\right)$

$\cos y=\cos (210^{o})$

We know that range of principal values is

$\cos$ is

$(0, \pi)$

$i.e (0^{o}, 180^{o})$

Hence

$4=210^{o}$ not possible

Now,

$\cos y=\cos (210^{o})$

$\cos y=\cos (360-150^{o})$

$\cos y=\cos (150^{o})$

$\cos y=\cos \left(\dfrac{5\pi}{5}\right)$

$4=\dfrac{5\pi}{6}$

Which is in range of principal values of

$\cos ^{-1}$ i.e

$(0, \pi)$

$\therefore y=\dfrac{5\pi}{6}$

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%
$none\ of\ these$

Explanation

$\Rightarrow 3\tan^{-1}\dfrac{1}{2}+2\tan^{-1}\dfrac{1}{5}+\sin^{-1}\dfrac{142}{65\sqrt{5}}$

$\Rightarrow \left(\tan^{-1}\dfrac{1}{\alpha}+\tan^{-1}\dfrac{1}{\alpha}\right)+\tan^{-1}\dfrac{1}{\alpha}+\left(\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{5}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$

$\Rightarrow \left(\tan^{-1}\dfrac{4}{3}+\tan^{-1}\dfrac{1}{\alpha}\right)+\tan^{-1}\left(\dfrac{5}{12}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$

$\Rightarrow \tan^{-1}\left(\dfrac{11}{2}\right)+\tan^{-1}\left(\dfrac{5}{12}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$

$\Rightarrow \pi +\tan^{-1}\left(\dfrac{-142}{31}\right)+\tan^{-1}\left(\dfrac{142}{31}\right)$

$\Rightarrow \pi$

Important property used

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

for

$x > o, y > o$

$xy < 1$

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

for

$x > 0, y > 0, xy > 1$ .

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%
$\dfrac {\pi}{3}+0.011547$

Explanation

$\sin^{-1}\left(0.5+0.01\right)=\sin^{1}\left(0.51\right)$

$\therefore$ Approx value

$=f \left(x\right)+hf’\left(x\right)$

Here

$x=0.5$

$h=0.01$

$f \left(u\right)=\sin^{1}x$

$f‘ \left(u\right)=\dfrac{1}{\sqrt{1-x^{2}}}$

$\therefore f \left(0.5\right)=\sin^{-1}\left(0.5\right)=\frac{\pi}{6}$

$\therefore f’ \left(0.5\right)=\dfrac{1}{\sqrt{1-\left(0.5\right)^{2}}}$

$=\dfrac{1}{\sqrt{0.75}}$

$=\boxed{\dfrac{2}{\sqrt{3}}}$

$\therefore$ Approx value

$= \dfrac{\pi}{6}+ \dfrac{2}{\sqrt{3}} \times \left(\dfrac{1}{100}\right)$

$\dfrac{\pi}{6}+\dfrac{2}{100\sqrt{3}}$

$\boxed{ =\dfrac{\pi}{6}+0.011541}$

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

$\left( 0<x<\dfrac { \pi }{ 2 } \right)$

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

Explanation

We have,

$\Rightarrow \cot^{-1}\left(\dfrac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right)$

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

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

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

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

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

$\Rightarrow \cot^{-1}\left(\dfrac{\cos x/2}{\sin x/2}\right)$

$\Rightarrow \cot^{-1}(-\cot x/2)$

$\Rightarrow \cot^{-1}(\cot (\pi -x/2))$

$\Rightarrow \pi -x/2$ .

1194705_1158127_ans_91afce2285eb46b0a7aa2f8d4738447f.jpg

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%
$0$

Explanation

$\tan^{-1}\left(\dfrac{1}{2}\right)+\tan^{-1}\left(\dfrac{1}{3}\right)$

use formula

$\tan^{-1}{a}+\tan^{-1}{b}$ =

$\tan^{-1} (\dfrac{a+b}{1-ab})$

$\Rightarrow \tan^{-1}\left(\dfrac{5}{6-1}\right)=\tan^{-1}\left(\dfrac{5}{5}\right)$

$=\tan^{1}(1)$

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

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

Explanation

$\Delta ABC$ ,

$\angle A=90^{\circ}$

So, from Pythagorus theorem,

$a^{2}=b^{2}+c^{2}$ .............. (1)

$\tan^{-1}\cfrac{b}{a+c}+\tan^{-1}\cfrac{c}{a+b}$

$=\tan^{-1}\cfrac{\cfrac{b}{a+c}+\cfrac{c}{a+b}}{1-\cfrac{b}{a+c}\cdot \cfrac{c}{a+b}}$

$=\tan^{-1}\cfrac{ab+b^{2}+ac+c^{2}}{a^{2}+ab+ac}$

$=\tan^{-1}\cfrac{ab+ac+a^{2}}{a^{2}+ab+ac}$ from equation (1)

$=\tan^{-1}(1)$

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

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

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

Explanation

Given

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

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

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

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

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

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

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

$\pi-\dfrac \pi 2$

$\implies \dfrac \pi 2$

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

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

Explanation

Given,

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

or,

${\tan ^{ - 1}}\dfrac{1}{{a - 1}} = {\tan ^{ - 1}}\dfrac{\dfrac{1}{x}+\dfrac{1}{a^2-x+1}}{1-\dfrac{1}{x(a^2-x+1)}}$

or,

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

or,

$(a-1)(a^2+1)=a^2 x-x^2+x-1$

or,

$a^3+a-a^2-1=a^2 x-x^2+x-1$

or,

$a^2(a-x)-(a^2-x^2)+(a-x)=0$

or,

$(a-x)(a^2-a-x+1)=0$

or,

$x=a, a^2-a+1$ .

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

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

Explanation

Solution :- given

$\displaystyle \Rightarrow cot^{-1}(2+\sqrt{3})$

$\displaystyle \Rightarrow cot^{-1}(\frac{2(2+\sqrt{3})}{2})$ multiply num and den by 2

$\displaystyle \Rightarrow cot^{-1}(\frac{4+2\sqrt{3}}{3-1})$

$\displaystyle \Rightarrow cot^{-1}(\frac{3+1+2\sqrt{3}}{(\sqrt{3})^{2}-(1)^{2}})$

$\displaystyle \Rightarrow cot^{-1}(\frac{(\sqrt{3})^{2}+(1)^{2}+2.1.\sqrt{3}}{(\sqrt{3}+1)(\sqrt{3}-1)})$

$\displaystyle \Rightarrow cot^{-1}(\frac{(\sqrt{3}+1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)})$

$\displaystyle \Rightarrow cot^{-1}(\frac{\sqrt{3}+1}{\sqrt{3}-1})$

$\displaystyle \Rightarrow cot^{-1}(\frac{(cot45\,cot30+1)}{cot30-cot45})$

$\displaystyle = cot^{-1} (cot (45-30))$

$= 45^{\circ}-30^{\circ}$

$\displaystyle = 15^{\circ}$

$\displaystyle = \frac{\pi }{12}$ Ans

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%
$- \dfrac{\pi }{{10}}$

Explanation

$\sin^{-1}\left(\cos\dfrac{33\pi}{5}\right)=\sin^{-1}\left(\cos\left(6\pi+\dfrac{3\pi}{5}\right)\right)$

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

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

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

$\therefore \sin^{-1}\left(\cos\dfrac{33\pi}{5}\right)=\dfrac{-\pi}{10}$

The value of

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

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

Explanation

$=\cos ( \tan^{-1} (\tan 2 ) ) = \cos 2$

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

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

Explanation

$\cot^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha})=x$

$\dfrac{\pi}{2}-\tan^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha})=x$ (because

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

$\Rightarrow \dfrac{\pi}{2}-2\tan^{-1}(\sqrt{\cos \alpha})=x$

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

Let's say

$\tan^{-1}(\sqrt{\cos \alpha})=\theta$

$\Rightarrow \tan\theta =\sqrt{\cos\alpha}$

$\sin x=sin\left(\dfrac{\pi}{2}-2\theta\right)=\cos 2\theta$

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

$=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

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

$=\dfrac{2\sin^2\alpha/2}{2\cos^2\alpha/2}$

$\Rightarrow \sin x=\tan^2\dfrac{\alpha}{2}$ .

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

$x$ is equal to:

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

Explanation

$\cos^{-1} \dfrac{3}{5} - \sin^{-1} \dfrac{4}{5} = \cos^{-1} x$

$\Rightarrow \cos^{-1} \dfrac{3}{5}- \cos^{-1} \dfrac{3}{5} = \cos^{-1} x$

$[\because \sin^{-1} \dfrac{4}{5} \Rightarrow \cos \theta = \dfrac{3}{5}]$

$\Rightarrow 0 = \cos ^{-1} x$

$\Rightarrow \cos 0 =x$

$\Rightarrow 1= x$

$\sin ^{ -1 }{ x } =\cot ^{ -1 }{ x }$ 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%
$\dfrac{{{\pi ^2}}}{4}\,and\,\dfrac{{ - \pi }}{4}$

Explanation

$\begin{aligned} \text { let } & \sin ^ { - 1 } x = 0 \\ & = x = \sin \theta \end{aligned}$

$\cos ^ { - 1 } x$ =

$\pi/2 - {\theta}$

thus

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

$o ^ { 2 } + \left( \frac { \pi } { 2 } - 0 \right) ^ { 2 }$

thus max value is =

$\frac { 5 \pi ^ { 2 } } { 4 }$

min value is =

$\frac { \pi ^ { 2 } } { 8 }$

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%
$4$

Explanation

$y=\sin^{-1}x+\cos^{-1}x+\sec^{-1}x$

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

$\sec^{-1}x\rightarrow x\in(-\infty, -1)\cup (1, \infty)$

$\therefore x=-1, 1$

$y(-1)=\dfrac{\pi}{2}+\sec^{-1}(-1)=\dfrac{\pi}{2}+\pi =\dfrac{3\pi}{2}$

$y(1)=\dfrac{\pi}{2}+\sec^{-1}1=\dfrac{\pi}{2}+0=\dfrac{\pi}{2}$

Hence, range contains two elements.

1098403_1196600_ans_43e07cd95e4a4c53be188546bdfcd08c.jpg

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

Explanation

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

$\because \tan^{-1}\dfrac{x}{2}=\cos^{-1}\dfrac{2}{\sqrt{4+x^2}}$

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

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

$\because \sin^{-1}\dfrac{2}{\sqrt{4+x^2}}=\tan^{-1}\dfrac{x}{2}$

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

$=\dfrac{2}{x}.$

Hence, the answer is

$\dfrac{2}{x}.$

1187518_1218543_ans_7a3d24dce5154310b266af62b854c2d3.png

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

$(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})$ is

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

Explanation

We know that the range of

${\sin}^{-1}{x}$ is

$\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$

Given:

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

$\Rightarrow\,{\sin}^{-1}{x}+{\sin}^{-1}{y}+{\sin}^{-1}{z}=\dfrac{\pi}{2}+\dfrac{\pi}{2}+\dfrac{\pi}{2}$

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

$\Rightarrow\,x=\sin{\dfrac{\pi}{2}},\,\,y=\sin{\dfrac{\pi}{2}},\,\,z=\sin{\dfrac{\pi}{2}}$

$\Rightarrow\,x=1,y=1,z=1$

$\left({x}^{500}+{y}^{500}+{z}^{500}\right)-\left({x}^{501}+{y}^{501}+{z}^{501}\right)$

$=\left({1}^{500}+{1}^{500}+{1}^{500}\right)-\left({1}^{501}+{1}^{501}+{1}^{501}\right)$

$=\left(1+1+1\right)-\left(1+1+1\right)=3-3=0$

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%
$\dfrac {1}{5}$

Explanation

$\sin \left(\dfrac 14 \sin ^{-1} \dfrac {\sqrt {63}}8\right)$

Let

$x=\sin ^{-1} \dfrac {\sqrt {63}}8$

$\implies \sin x=\dfrac {\sqrt {63}}8$

$\implies \cos x=\dfrac {\sqrt {64-63}}{8}=\dfrac 18$

$\cos \dfrac x2 =\sqrt {\dfrac {1+\cos x}{2}}$

$=\sqrt {\dfrac 12 \left(\dfrac 98\right)}$

$\cos \dfrac x2=\dfrac 34$

$\implies \cos \dfrac x4 =\sqrt {\dfrac {1+\cos \dfrac x2}2}$

$=\sqrt {\dfrac 12 \left(1+\dfrac 34\right)}$

$=\sqrt{\dfrac 78}$

$\implies \sin \dfrac x4$

$=\sqrt{1-\cos ^2\dfrac x4}$

$=\sqrt {1-\dfrac 78}$

$=\dfrac 1{2\sqrt 2}$

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

Explanation

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

We know that, range of

$\cos^{-1}x$ is

$[0,\pi]$ and range of

$\cos^{-1}2x$ is

$[0,\pi]$ .

So, here we can see both ranges are in positive quantity.

So, sum of positive quantity cannot be negative quantity.

$\therefore$

$\cos ^ { - 1 } x + \cos ^ { - 1 } 2 x = - \pi$ is not possible.

$\therefore$ The number of real solution is

$0.$

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

Explanation

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

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

$\Rightarrow \cos^{-1}\left(\cos\left(2\pi +\dfrac{7\pi}{5}\right)\right)$

$\Rightarrow \cos^{-1}\left(\cos\dfrac{7\pi}{5}\right)$

$\Rightarrow \cos^{-1}\left(\cos\left(2\pi -\dfrac{3\pi}{5}\right)\right)$

$=\cos^{-1}\cos \dfrac{3\pi}{5}$

$=\dfrac{3\pi}{5}$ .

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

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

Explanation

We have,

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

Let

$\cos h^{-1}x=A--------------(1)$

$\cos h\ A=x$

Squaring both side

$\cos h^{2} A=x^{2}$

$1+\sin h^{2} A=x^{2}$

$\because \cos h^{2}x- \sin h^{2} x=1$

$\sin h^{2} A=x^{2}-1$

$\sin h\ A= \sqrt{x^{2}-1}$

$A= \sin h^{-1} \sqrt{x^{2}-1} ---------(2)$

By equation

$(1)$ and

$(2)$ to

$\cos h^{-1} x= \sin h^{-1} \sqrt{x^{2}-1}$

According to given function.

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

$\Rightarrow \sin h \sin h^{-1} \sqrt{x^{2}-1}$

$\Rightarrow \sqrt{x^{2}-1}$

Hence, this is the answer.

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

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

Explanation

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

Now, we know that

$\sin h^{-1}x=\log{(x+\sqrt{x^2+1})}$

then,

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

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

$=\log{(\sqrt{8}+\sqrt{9})}$

$=\log{(\sqrt{9}+\sqrt{8})}$

$=\log{(\sqrt{3^2}+\sqrt{8})}$

$\sin h^{-1}(2^{3/2})=\log{(3+\sqrt{8})}$

Hence, this is the answer.

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

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

Explanation

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

$= \tan^{-1} \left(\dfrac{2 - 1}{1 + 2.1} \right) + \tan^{-1} \left(\dfrac{4 - 2}{1 + 4.2} \right) + \tan^{-1} \left(\dfrac{8 - 4}{1 + 8.4} \right) + \tan^{-1} \left(\dfrac{16 - 8}{1 + 16 \times 8} \right) +...+ \tan^{-1} \left(\dfrac{2^n - 2^{n - 1}}{1 + 2^n \times 2^{n - 1}} \right)$

we know that

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

$= \tan^{-1} 2 - \tan^{-1} 1 + \tan^{-1} 4 - \tan^{-1} 2 + \tan^{-1} 8 - \tan^{-1} 4 + \tan^{-1} 16 - \tan^{-1} 8 + ... + \tan^{-1} 2^n - \tan^{-1} 2^{n - 1}$

$= \tan^{-1} 2^n - \tan^{-1} 1$

$= \tan^{-1} 2^n - \dfrac{\pi}{4}$

$\cot^{-1} [\sqrt{\cos \alpha}] - \tan^{-1} [\sqrt{\cos \alpha}] = x$ , then

$\sin x$ is equal to

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

Explanation

Given equation:

${ cot }^{ -1 }(\sqrt { cos\alpha } )-{ tan }^{ -1 }(\sqrt { cos\alpha } )=x$

$let\quad { cot }^{ -1 }(\sqrt { cos\alpha } )=P$

$\Longrightarrow cotP=\sqrt { cos\alpha }$

$\Longrightarrow sinP=\dfrac { 1 }{ \sqrt { 1+cos\alpha } }$

Also

${ tan }^{ -1 }(\sqrt { cos\alpha } )=Q$

$\Longrightarrow tanQ=\sqrt { cos\alpha }$

$\Longrightarrow sinQ=\dfrac { \sqrt { cos\alpha } }{ \sqrt { 1+cos\alpha } }$

Now the given equation can be written as

$x=P-Q$

$sinx=sin(P-Q)$

$sinx=sinPcosQ-cosPsinQ$

$sinx=\left( \dfrac { 1 }{ \sqrt { 1+cos\alpha } } \right) \left( \dfrac { 1 }{ \sqrt { 1+cos\alpha } } \right) -\left( \dfrac { \sqrt { cos\alpha } }{ \sqrt { 1+cos\alpha } } \right) \left( \dfrac { \sqrt { cos\alpha } }{ \sqrt { 1+cos\alpha } } \right)$

$sinx=\dfrac { 1-cos\alpha }{ 1+cos\alpha } =\dfrac { 2{ sin }^{ 2 }\dfrac { \alpha }{ 2 } }{ 2{ cos }^{ 2 }\dfrac { \alpha }{ 2 } } \\ sinx={ tan }^{ 2 }\dfrac { \alpha }{ 2 }$

Hence option (a) is correct.

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

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

Explanation

We have,

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

We know that,

$\sec h^{-1}x=\log \left (\dfrac {1}{x}+\sqrt {\dfrac {1}{x^{2}}1}\right)$

Now,

$\sec ^{ -1 }{ \dfrac { 1 }{ 5 } } =\log { \left( \dfrac { 1 }{ \dfrac { 1 }{ 5 } } +\sqrt { \dfrac { 1 }{ { \left( \dfrac { 1 }{ 5 } \right) }^{ 2 } } -1 } \right) }$

$=\log { \left( 5+\sqrt { 25-1 } \right) }$

$=\log { \left( 5+\sqrt { 24 } \right) }$

Hence this is the answer.

$\cot \dfrac {2x}{3}+\tan \dfrac {x}{3}=\csc \dfrac {x}{3}$ then value of

$\tan^{-1(\tan k)}$ equals

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

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

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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