Question

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

• A. $\frac{4}{5}$
• B. $-\frac{4}{5}$
• C. $\frac{3}{5}$
• D. $0$

Answer 1

The correct option is B $-\frac{4}{5}$

Now

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

$se{c}^{-1}(-2)=\frac{2\pi }{3}$

$cose{c}^{-1}(-\sqrt{2})=\frac{-\pi }{4}$

$co{s}^{-1}(-0.5)=\frac{2\pi }{3}$

So putting the values, we have

$\frac{\frac{\pi }{3}-\frac{2\pi }{3}}{\frac{-\pi }{4}+\frac{2\pi }{3}}$

$=-\frac{4}{5}$

answer(b)