Question:
The value of $\tan \left\{\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right\}$ is
(a) $\frac{\sqrt{29}}{3}$
(b) $\frac{29}{3}$
(c) $\frac{\sqrt{3}}{29}$
(d) $\frac{3}{29}$
Solution:
(d) $\frac{3}{29}$
Let, $\cos ^{-1} \frac{1}{5 \sqrt{2}}=y$ and $\sin ^{-1} \frac{4}{\sqrt{17}}=z$
$\therefore \cos y=\frac{1}{5 \sqrt{2}} \Rightarrow \sin y=\frac{7}{5 \sqrt{2}} \Rightarrow \tan y=7$
$\sin z=\frac{4}{\sqrt{17}} \Rightarrow \cos z=\frac{1}{\sqrt{17}} \Rightarrow \tan z=4$
$\therefore \tan \left(\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right)=\tan (y-z)$
$=\frac{\tan y-\tan z}{1+\tan y \tan z}$
$=\frac{7-4}{1+7 \times 4}$
$=\frac{3}{29}$