Question

$\frac{ta{n}^{-1}\left(\sqrt{3}\right)-se{c}^{-1}\left(-2\right)}{cose{c}^{-1}\left(-\sqrt{2}\right)+co{s}^{-1}\left(-\frac{1}{2}\right)}$ is equal to

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

Answer 1

Answer: (B)

$\frac{{\tan }^{-1}(\sqrt{3})-{\sec }^{-1}(-2)}{{\mathrm{cosec}}^{-1}(-\sqrt{2})+{\cos }^{-1}\left(-\frac{1}{2}\right)}$
$=\frac{{\tan }^{-1}(\sqrt{3})-\left(\pi -{\sec }^{-1}(2)\right)}{-{\mathrm{cosec}}^{-1}(\sqrt{2})+\pi -{\cos }^{-1}\left(\frac{1}{2}\right)}$
$=\frac{\frac{\pi }{3}-\left(\pi -\frac{\pi }{3}\right)}{-\frac{\pi }{4}+\pi -\frac{\pi }{3}}=\frac{\frac{2\pi }{3}-\pi }{\pi -\frac{\pi }{4}-\frac{\pi }{3}}$
$=\frac{\frac{-\frac{\pi }{3}}{12\pi -3\pi -4\pi }}{12}=\frac{-\frac{\pi }{3}}{\frac{5\pi }{12}}=-\frac{4}{5}$