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UPSC NDA I 2015 Mathematics Solved Paper

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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 : 42 of 120

Marks: +1, -0

what is

x-y+z

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

Solution:

From the above question, we obtained

x-y

=\tan^{-1}\left(\frac{120}{119}\right)-\tan^{-1}\left(\frac{1}{70}\right)

=\tan^{-1}\left(\frac{\frac{120}{119}-\frac{1}{70}}{1+\frac{120}{70\times 19}}\right)

=\tan^{-1}\left(\frac{8281}{8450}\right)

=\tan^{-1}\left(\frac{49}{50}\right)

Therefore

x-y+z

=\tan^{-1}\left(\frac{49}{50}\right)+\tan^{-1}\left(\frac{1}{99}\right)

=\tan^{-1}\left(\frac{\frac{49}{50}+\frac{1}{99}}{1-\frac{49}{50\times 99}}\right)

=\tan^{-1}\left(\frac{4901}{4901}\right)

=\tan^{-1}(1)

=\frac{\pi}{4}

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