Question

The value of $\left({\tan }^{-1}\pi +{\tan }^{-1}\left(\frac{1}{\pi }\right)\right)+{\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)$ is equal to

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

Answer 1

The correct option is B $\frac{\pi }{6}$

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

$={\tan }^{-1}\pi +{\cot }^{-1}\pi +{\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)$ $[\because {\tan }^{-1}(\frac{1}{y})={\cot }^{-1}y]$

$=\frac{\pi }{2}+{\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)$ $[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}]$

$=\frac{\pi }{2}+\frac{\pi }{3}-\frac{2\pi }{3}$ $[\because \tan \frac{\pi }{3}=\sqrt{3};\sec \frac{2\pi }{3}=-2]$

$=\frac{\pi }{2}-\frac{\pi }{3}$

$=\frac{\pi }{6}$

None of the given options are correct.