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Inverse Trigonometric Functions

Section: Mathematics

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Question : 77 of 99

Marks: +1, -0

\csc[2 \cot^{-1}(5)+\cos^{-1}\left(\;\frac{4}{5}\right)]

[25 Feb 2021 Shift 2]

$\;\frac{56}{33}$
$\;\frac{65}{33}$
$\;\frac{65}{56}$
$\;\frac{75}{56}$

Solution:

\csc[2 \cot^{-1}(5)+\cos^{-1}\left(\frac{4}{5}\right)]

=\csc[2 \tan^{-1}\left(\;\frac{1}{5}\right)+\cos^{-1}\left(\;\frac{4}{5}\right)] \;\; \left[ \because \tan^{-1} x = \cot^{-1}\left(\;\frac{1}{x}\right) \right]

=\csc[\tan^{-1}\left(\; \frac{2\left(\frac{1}{5}\right)}{1-\left(\frac{1}{5}\right)^{2}}\right)+\cos^{-1}\left(\;\frac{4}{5}\right)] \left[\therefore 2\tan^{-1}\theta=\tan^{-1}\left(\;\frac{2\theta}{1-\theta^{2}}\right)\right]

=\csc\left(\tan^{-1} \;\frac{5}{12} + \cos^{-1} \;\frac{4}{5}\right)

Let

\tan^{-1}\left(\;\frac{5}{12}\right)=x

, then

\tan x = \frac{5}{12}

gives

\sin x = \frac{5}{13}, \cos x = \frac{12}{13}

Let

\cos^{-1}\left(\;\frac{4}{5}\right) = y

, then

\cos y = \frac{4}{5}

gives,

\sin y = \frac{3}{5}

Now,

\csc(x+y) = \;\frac{1}{\sin(x+y)} = \;\frac{1}{\sin x \cos y + \cos x \sin y}

= \;\frac{1}{\left(\;\frac{5}{13}\right)\left(\;\frac{4}{5}\right) + \left(\;\frac{12}{13}\right)\left(\;\frac{3}{5}\right)} = \;\frac{65}{56}

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