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UPSC NDA Math Model Paper 10

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Question : 68 of 120

Marks: +1, -0

What is the value of

\sin^{-1} \frac{4}{5}+2 \tan^{-1} \frac{1}{3}

?

$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$

Solution:

Let,

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

\sin x = \frac{4}{5}

Now,

\cos^2 x = 1-\sin^2 x

=1-\frac{16}{25}=\frac{9}{25}

⇒ cos x = 3/5

Now, cot x = cos x/sin x = 3/4 ....(1)

2 \tan^{-1} \frac{1}{3} = \tan^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right)

\text{.......} \left(\because 2 \tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right)

=\tan^{-1}\left(\frac{\frac{3}{3}}{\frac{8}{9}}\right)

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

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

....(from 1)

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

=\frac{\pi}{2}-x

\text{.......} .(\because \tan^{-1}(\tan x)=x)

=\frac{\pi}{2}-\sin^{-1} \frac{4}{5}

\therefore \sin^{-1} \frac{4}{5}+2 \tan^{-1} \frac{1}{3}=\sin^{-1} \frac{4}{5}

+\frac{\pi}{2}-\sin^{-1} \frac{4}{5}

=\frac{\pi}{2}

Hence, option (2) is correct.

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