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NDA Math Inverse Circular Functions

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For the next three (3) items that follow:

Consider

x=4\tan^{-1}\left(\frac{1}{5}\right),y=\tan^{-1}\left(\frac{1}{70}\right)

and

z=\tan^{-1}\left(\frac{1}{99}\right)

[NDA I 2015]

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Question : 19 of 25

Marks: +1, -0

What is x equal to ?

$\tan^{-1}\left(\frac{60}{119}\right)$
$\tan^{-1}\left(\frac{120}{119}\right)$
$\tan^{-1}\left(\frac{90}{169}\right)$
$\tan^{-1}\left(\frac{170}{169}\right)$

Solution:

Applying trigonometric identities yields

x=4\tan^{-1}\left(\frac{1}{5}\right)

\frac{x}{4}=\tan^{-1}\left(\frac{1}{5}\right)

\tan\left(\frac{x}{4}\right)=\frac{1}{5}

\therefore\tan\left(\frac{x}{2}\right)=\frac{2\tan\left(\frac{x}{4}\right)}{1-\tan^2\left(\frac{x}{4}\right)}

=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{5}{12}

\therefore\tan\left(\frac{x}{2}\right)=\frac{5}{12}

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

=\frac{\frac{10}{12}}{1-\frac{25}{144}}=\frac{120}{119}

\therefore\tan(x)=\frac{120}{119}

x=\tan^{-1}\left(\frac{120}{119}\right)

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