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  4. tan( cos-1( (1/5√2))- sin-1((4/√17))) is

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Q. $\tan\left(\cos^{-1}\left( \frac{1}{5\sqrt{2}}\right)-\sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right)$ is

COMEDKCOMEDK 2015Inverse Trigonometric Functions

Solution:

$ \tan\left(\cos^{-1}\left( \frac{1}{5\sqrt{2}}\right)-\sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right)$
$= \tan\left(\frac{\pi}{2} - \sin^{-1}\left( \frac{1}{5\sqrt{2}}\right)-\sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right)$
$= \tan\left(\frac{\pi}{2} - \left(\sin^{-1} \frac{4}{\sqrt{17}}+\sin^{-1}\frac{1}{5\sqrt{2}}\right)\right)$
$= \tan\left(\frac{\pi}{2} - \sin^{-1} \left(\frac{4}{\sqrt{17}} \sqrt{1-\frac{1}{50} }+\frac{1}{5\sqrt{2}} \sqrt{1-\frac{16}{17}}\right)\right)$
$= \tan\left(\frac{\pi}{2} - \sin^{-1} \left(\frac{4}{\sqrt{17}} . \frac{7}{5\sqrt{2}} +\frac{1}{5\sqrt{2}}. \frac{1}{\sqrt{17}}\right)\right)$
$=\cot\left(\sin^{-1} \frac{29}{5\sqrt{2}.\sqrt{17}}\right)=\cot\left(\cot^{-1} \frac{3}{29}\right)$
$ = \frac{3}{29}$.