Question

${\sin }^{−1}\left(\sin \frac{2π}{3}\right)+{\cos }^{−1}\left(\cos \frac{7π}{6}\right)+{\tan }^{−1}\left(\tan \frac{3π}{4}\right)$ is equal to :

• A. $\frac{11Ï€}{12}$
• B. $\frac{17Ï€}{12}$
• C. $\frac{31Ï€}{12}$
• D. $−\frac{3Ï€}{4}$

Answer 1

Answer: (A)

${\sin }^{−1}\left(\sin \frac{2π}{3}\right)+{\cos }^{−1}\left(\cos \frac{7π}{6}\right)+{\tan }^{−1}\tan \left(\frac{3π}{4}\right)$
${\sin }^{−1}\sin \left(\frac{2π}{3}\right)=π−\frac{2π}{3}=\frac{π}{3}$
${\cos }^{−1}\left(\cos \frac{2π}{6}\right)=2π−\frac{7π}{6}=\frac{5π}{6}$
${\tan }^{−1}\tan \left(\frac{3π}{4}\right)=\frac{3π}{4}−π=\frac{−π}{4}$
${\sin }^{−1}\left(\sin \frac{2π}{3}\right)+{\cos }^{−1}\cos \frac{7π}{6}+{\tan }^{−1}\tan \frac{3π}{4}$
$=\frac{11Ï€}{12}$