Mathematics M.C.Q (1 Marks)

==> ${\tan ^{ - 1}}(1 + x) = \frac{\pi }{2} - {\tan ^{ - 1}}(1 - x)$

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

==> ${\tan ^{ - 1}}(1 + x) = {\tan ^{ - 1}}\left( {\frac{1}{{1 - x}}} \right)$

==> $1 + x = \frac{1}{{1 - x}} \Rightarrow 1 - {x^2} = 1 \Rightarrow x = 0$.

${\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$

==> ${\cot ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{2}$ = $\frac{\pi }{6}$ 

==> $x = \sqrt 3 $.

Hence given equation can be written as

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

$ \Rightarrow $ $x = \frac{1}{{\sqrt 5 }}$.

==> $3{\sin ^{ - 1}}x + {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi $

==> $3{\sin ^{ - 1}}x = \pi - \frac{\pi }{2} = \frac{\pi }{2}$

==> ${\sin ^{ - 1}}x = \pi /6$

==> $x = \sin \frac{\pi }{6} = \frac{1}{2}$.

$ \Rightarrow C = \sin \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cos }^{ - 1}}\frac{{12}}{{13}}} \right)$

using, $\sin (A + B) = \sin A\,\,\cos B + \cos A\,\sin B$

$ \Rightarrow \,\,C = \frac{3}{5} \times \frac{{12}}{{13}} + \sqrt {1 - \frac{9}{{25}}} \sqrt {1 - \frac{{144}}{{169}}} $

$ \Rightarrow C = \frac{{56}}{{65}}$.

Putting $x = \tan \theta $ we get,

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

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

$= \sin [{\tan ^{ - 1}}\tan (\pi /2 - 2\theta ) + {\cos ^{ - 1}}\cos 2\theta ]$

$= \sin \frac{\pi }{2} = 1$.

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

$= \tan \left[ {{{\tan }^{ - 1}}\frac{{\frac{1}{2} - \frac{1}{3}}}{{1 + \frac{1}{6}}}} \right]$

$= \tan {\tan ^{ - 1}}\left( {\frac{1}{6} \times \frac{6}{7}} \right)$= $\frac{1}{7}$.

Now $x = \tan A \Rightarrow \sin 2A = \frac{{2\tan A}}{{1 + {{\tan }^2}A}} = \frac{{2x}}{{1 + {x^2}}}$.

$ \Rightarrow \cos ({\sin ^{ - 1}}2x\sqrt {1 - {x^2}} ) = \frac{1}{9}$

==>$\cos ({\cos ^{ - 1}}\sqrt {1 - 4{x^2} + 4{x^4}} ) = \frac{1}{9}$

==> $1 - 2{x^2} = \frac{1}{9} \Rightarrow 2{x^2} = 1 - \frac{1}{9} = \frac{8}{9}$

==> ${x^2} = \frac{4}{9} \Rightarrow x = \pm \frac{2}{3}$.

==> ${\cos ^{ - 1}}\sqrt {\left( {\frac{{1 + x}}{2}} \right)} = \frac{\pi }{4} $

$\Rightarrow \cos \frac{\pi }{4} = \frac{{\sqrt {1 + x} }}{{\sqrt 2 }}$

==> $\frac{1}{{\sqrt 2 }} = \frac{{\sqrt {1 + x} }}{{\sqrt 2 }} $

$\Rightarrow 1 = \sqrt {1 + x} \Rightarrow x = 0$.

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

$\frac{{x + y}}{{1 - xy}} = 1$;

$x + y + xy = 1$.

$ = \sin \left[ {{{\sin }^{ - 1}}\left\{ {3{\rm{ }}\left( {\frac{1}{5}} \right) - 4{\rm{ }}{{\left( {\frac{1}{5}} \right)}^3}} \right\}} \right]$

$ = \sin \left[ {{{\sin }^{ - 1}}\left\{ {\frac{3}{5} - \frac{4}{{125}}} \right\}} \right]$

$ = \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{{75 - 4}}{{125}}} \right)} \right]$

$ = \sin \left[ {{{\sin }^{ - 1}}\frac{{71}}{{125}}} \right] = \frac{{71}}{{125}}$.

$ \Rightarrow \tan \theta \tan \phi = - 1 \Rightarrow \tan \theta = - \cot \phi $

$ \Rightarrow \theta - \phi = \frac{\pi }{2}$.

Then, $\tan x=\sqrt{3}=\tan \frac{\pi}{3}$

We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ $\therefore \tan ^{-1} \sqrt{3}=\frac{\pi}{3}$

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

Then, $\sec y=-2=-\sec \left(\frac{\pi}{3}\right)=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \frac{2 \pi}{3}$

We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$

$\therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}$

Thus, $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$

$=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}$

$=\tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}\right]$

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

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

Therefore, $\cot ^{-1} \frac{1}{\sqrt{x^{2}-1}}=\cot ^{-1}(\cot \theta)$$=\theta=\sec ^{-1} x$

Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

$\therefore \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\tan ^{-1} \frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}$

$=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)$

$\tan ^{-1}\left(\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)$

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

Put $x=cosec \theta \Rightarrow \theta=cosec^{-1} x$

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

$=\tan ^{-1} \frac{1}{\sqrt{\cos e c^{2} \theta-1}}$

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

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

$=\theta$

$=cosec ^{-1} x$

$=\frac{\pi}{2}-\sec ^{-1} x $    $\left[A s, \cos e c^{-1} x+\sec ^{-1} x=\frac{\pi}{2}\right]$

$=\cot \left(\frac{\pi}{2}\right)$

$=0$

Then

$\sin x=\frac{3}{5}$

$\Rightarrow \cos x=\sqrt{1-\sin ^{2} x}=\frac{4}{5}$

$\Rightarrow \sec x=\frac{5}{4}$

$\therefore \tan x=\sqrt{\sec ^{2} x-1}=\sqrt{\frac{25}{16}-1}=\frac{3}{4}$

$\therefore x=\tan ^{-1} \frac{3}{4}$

$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$

Now, $\cot ^{-1} \frac{3}{2}=\tan ^{-1} \frac{2}{3}$

Therefore, $\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$

$=\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)$

$=\tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right]$

$=\tan \left(\tan ^{-1} \frac{9+8}{12-6}\right)$

$=\tan \left(\tan ^{-1} \frac{17}{6}\right)=\frac{17}{6}$

Then $, \sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)$

We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ $\sin ^{-1}\left(\frac{-1}{2}\right)=\frac{\pi}{6}$

$\therefore \sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)$$=\sin \left(\frac{3 \pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)=1$

Now, $\cos ec \theta =\sqrt {1+ {\cot ^2 \theta}} = \sqrt {{x^2} +1} $.

$\therefore \,\,\,\sin \theta  = \frac{1}{{\cos ec\,\theta }} = \frac{1}{{\sqrt {1 + {x^2}} }}\,\, \Rightarrow \,\theta  = {\sin ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}$

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

$ = \frac{1}{{\sqrt {1 + {x^2}} }} = {(1 + {x^2})^{ - 1/2}}$

Now $1 + {\cot ^2}\theta = \cos e{c^2}\theta = \frac{1}{{{x^2}}}$

Hence $1 + {\cot ^2}\,({\sin ^{ - 1}}x) = \frac{1}{{{x^2}}}$.

$ = {\tan ^{ - 1}}(\tan \theta ) = \theta = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right)$.

$\therefore \,\,\cos \theta = \frac{1}{{\sqrt {1 + {{\tan }^2}\theta } }} = \frac{1}{{\sqrt {1 + {x^2}} }}$

Hence $\cos \theta = \cos \,({\tan ^{ - 1}}x) = \frac{1}{{\sqrt {1 + {x^2}} }}$.

$= {\cos ^{ - 1}}\left( { - \cos \frac{\pi }{6}} \right) = \pi - {\cos ^{ - 1}}\cos \frac{\pi }{6} $

$= \pi - \frac{\pi }{6} = \frac{{5\pi }}{6}$.

${\sin ^{ - 1}}\frac{{\sqrt x }}{{\sqrt {x + a} }} = {\sin ^{ - 1}}\frac{{\sqrt a \sqrt {{{\tan }^2}\theta } }}{{\sqrt {a\,{{\tan }^2}\theta + a} }} $

$= {\sin ^{ - 1}}\frac{{\sqrt a \,\tan \theta }}{{\sqrt a \,\sec \theta }}$

$ = {\sin ^{ - 1}}\sin \theta = \theta = {\tan ^{ - 1}}\left( {\sqrt {\frac{x}{a}} } \right)$.

$ = \frac{{|x|}}{{\sqrt {{x^2} - 1} }}$, for $|x|\, > \,1$.

$ = {\cos ^{ - 1}}\left[ {\cos \left( {2\pi - \frac{\pi }{3}} \right)} \right] + {\sin ^{ - 1}}\left[ {\sin \left( {2\pi - \frac{\pi }{3}} \right)} \right]$

$ = \frac{\pi }{3} - \frac{\pi }{3} = 0$.

Now $\sin \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{4}{5}} \right) = \sin \left( {\frac{x}{2}} \right)$ .....$(ii)$

From $(i),$ $\cos x = \frac{4}{5}$

==> $1 - 2{\sin ^2}\frac{x}{2} = \frac{4}{5}$

==> $2{\sin ^2}\frac{x}{2} = 1 - \frac{4}{5} = \frac{1}{5}$

$ \Rightarrow \sin \frac{x}{2} = \sqrt {\frac{1}{{10}}} $.

Then, $cosec y=2=cosec \left(\frac{\pi}{6}\right)$

We know that the range of the principal value branch of $cosec ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$.

Therefore, the principal value of $cosec ^{-1}(2)$ is $\frac{\pi}{6}$

Then, $\tan y=-\sqrt{3}=-\tan \frac{\pi}{3}=\tan \left(-\frac{\pi}{3}\right)$

We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2} \frac{\pi}{2}\right)$ and $\tan \left(-\frac{\pi}{3}\right)$ is $-\sqrt{3}$

Therefore, known that the principal value of $\tan ^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$

Then, $\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)$

We know that the range of the principal value branch of $\cos ^{-1}$ is $[0, \pi]$ and $\cos \left(\frac{2 \pi}{3}\right)=-\frac{1}{2}$

Therefore, the principal value of $\cos ^{-1}\left(-\frac{1}{2}\right)$ is $\left(\frac{2 \pi}{3}\right)$

Then, $\tan y=-1=-\tan \left(\frac{\pi}{4}\right)=\tan \left(-\frac{\pi}{4}\right)$

We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\tan \left(-\frac{\pi}{4}\right)=-1$

Therefore, the principal value of $\tan ^{-1}(-1)$ is $-\frac{\pi}{4}$

We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$ and $\sec \left(\frac{\pi}{6}\right)=\frac{2}{\sqrt{3}}$

Therefore, the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$

We know that the range of the principal value branch of $\cot ^{-1}$ is $(0, \pi)$ and $\cot \left(\frac{\pi}{6}\right)=\sqrt{3}$

Therefore, the principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$

Then $\cos y=-\frac{1}{\sqrt{2}}=-\cos \left(\frac{\pi}{4}\right)=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \left(\frac{3 \pi}{4}\right)$

We know that the range of the principal value branch of $\cos ^{-1}$ is $[0, \pi]$ and $\cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}}$

Therefore, the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$

We know that the range of the principal value branch of cosec $^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$ and $cosec \left(-\frac{\pi}{4}\right)=-\sqrt{2}$

Therefore, the principal value of $cosec ^{-1}(-\sqrt{2})$ is $-\frac{\pi}{4}$

Then, $\tan x=1=\tan \left(\frac{\pi}{4}\right)$

$\therefore \tan ^{-1}(1)=\frac{\pi}{4}$

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

Then, $\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)$

$\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}$

Let $\sin ^{-1}\left(-\frac{1}{2}\right)=z$

Then, $\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)$

$\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$

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

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

$=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}$

We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

Let $\tan ^{-1} x=y .$ Then

$y=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$ $\Rightarrow \tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$

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

Putting $a = \tan \theta $ and $b = \tan \phi $

So, ${\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + {\sin ^{ - 1}}\left( {\frac{{2\tan \phi }}{{1 + {{\tan }^2}\phi }}} \right) = 2{\tan ^{ - 1}}x$

==> ${\sin ^{ - 1}}\sin (2\theta ) + {\sin ^{ - 1}}\sin (2\phi ) = 2{\tan ^{ - 1}}x$

==> $2(\theta + \phi ) = 2{\tan ^{ - 1}}x$

Hence $x = \tan (\theta + \phi )$

==> $x = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }}$

Substituting these values, we get $x = \frac{{a + b}}{{1 - ab}}$.

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

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

$ = {\tan ^{ - 1}}\frac{x}{y} + {\cot ^{ - 1}}\frac{x}{y} - \frac{\pi }{4} = \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}$.

==> ${({\tan ^{ - 1}}x + {\cot ^{ - 1}}x)^2} - 2{\tan ^{ - 1}}x\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}x} \right) = \frac{{5{\pi ^2}}}{8}$

==> $\frac{{{\pi ^2}}}{4} - 2 \times \frac{\pi }{2}{\tan ^{ - 1}}x + 2{({\tan ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$

==> $2{({\tan ^{ - 1}}x)^2} - \pi {\tan ^{ - 1}}x - \frac{{3{\pi ^2}}}{8} = 0$

==> ${\tan ^{ - 1}}x = - \frac{\pi }{4},\frac{{3\pi }}{4}$

==> ${\tan ^{ - 1}}x = - \frac{\pi }{4} \Rightarrow x = - 1$.