Question

$-\frac{2 \pi}{5}$ is the principal value of

• A. $\cos ^{-1}\left(\cos \frac{7 \pi}{5}\right)$
• B. $\sin ^{-1}\left(\sin \frac{7 \pi}{5}\right)$
• C. $\sec ^{-1}\left(\sec \frac{7 \pi}{5}\right)$
• D. $\sin ^{-1}\left\{\sin \left(\frac{2 \pi}{5}\right)\right\}$

Answer 1

Answer: (B)

Since, $\frac{7 \pi}{5}$ lies in III quadrant
$\therefore \sin \left(\frac{7 \pi}{5}\right)<0$ and $\cos \left(\frac{7 \pi}{5}\right)<0$
Then, $\sec \left(\frac{7 \pi}{5}\right)<0$
$-\frac{\pi}{2} \leq \sin ^{-1}\left(\sin \frac{7 \pi}{5}\right)<0$
and $\frac{\pi}{2}<\cos ^{-1}\left(\cos \frac{7 \pi}{5}\right) \leq \pi$
and $\frac{\pi}{2}<\sec ^{-1}\left\{\sec \frac{7 \pi}{5}\right\} \leq \pi$
Hence, alternate (a) and (c) are fail
Now, $\sin ^{-1}\left(\sin \frac{7 \pi}{5}\right)=\sin ^{-1}\left(\sin \left(\pi+\frac{2 \pi}{5}\right)\right)$
$=\sin ^{-1}\left(-\sin \frac{2 \pi}{5}\right)$
$=\sin ^{-1}\left\{\sin \left(-\frac{2 \pi}{5}\right)\right\}=-\frac{2 \pi}{5}$