Question

${\tan }^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+{\sec }^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$ is equal to :

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{2}$.
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (D)

${\tan }^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+{\sec }^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$
$={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi }{3}$