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  4. The value of cos (2 sin -1((-1/6)))+ tan ( sec -1((-13/12))) is

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Q. The value of $\cos \left(2 \sin ^{-1}\left(\frac{-1}{6}\right)\right)+\tan \left(\sec ^{-1}\left(\frac{-13}{12}\right)\right)$ is

Inverse Trigonometric Functions

Solution:

$\sin ^{-1}\left(\frac{-1}{6}\right)=\theta \Rightarrow \sin \theta=\frac{-1}{6}$
So, $\cos \left(2 \sin ^{-1}\left(\frac{-1}{6}\right)\right)=\cos 2 \theta=1-2 \sin ^2 \theta=1-\frac{2}{36}=\frac{17}{18}$
Also, let $\sec ^{-1}\left(\frac{-13}{12}\right)=\theta \Rightarrow \sec \theta=\frac{-13}{12}\left(2^{\text {nd }}\right.$ quadrant $) \Rightarrow \tan \theta=\frac{-5}{12}$
$\therefore$ We have, $\frac{17}{18}-\frac{5}{12}=\frac{34-15}{36}=\frac{19}{36}$. Ans.]