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, $\sec \left(\frac{7\pi }{5}\right)<0$
$-\frac{\pi }{2}\le {\sin }^{-1}\left(\sin \frac{7\pi }{5}\right)<0$
and $\frac{\pi }{2}<{\cos }^{-1}\left(\cos \frac{7\pi }{5}\right)\le \pi$
and $\frac{\pi }{2}<{\sec }^{-1}\left\{\sec \frac{7\pi }{5}\right\}\le \pi$
Hence, alternate (a) and (c) are fail
Now, ${\sin }^{-1}\left(\sin \frac{7\pi }{5}\right)={\sin }^{-1}\left(\sin \left(\pi +\frac{2\pi }{5}\right)\right)$
$={\sin }^{-1}\left(-\sin \frac{2\pi }{5}\right)$
$={\sin }^{-1}\left\{\sin \left(-\frac{2\pi }{5}\right)\right\}=-\frac{2\pi }{5}$