Question

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

• A. $\frac{29}{36}$
• B. $\frac{19}{36}$
• C. $\frac{7}{36}$
• D. $\frac{49}{36}$

Answer 1

Answer: (B)

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