Question

$ta{n}^{-1}\sqrt{3}-co{t}^{-1}(-\sqrt{3})$ is equal to
a) $\pi$
b) $-\frac{\pi }{2}$
c) zero
d) $2\sqrt{3}$

Answer 1

$\begin{array}{cc}ta{n}^{-1}\sqrt{3}-co{t}^{-1}(-\sqrt{3})& \\ =ta{n}^{-1}\sqrt{3}-[\pi -co{t}^{-1}\sqrt{3}]& [\because co{t}^{-1}(-x)=\pi -co{t}^{-1}x]\\ =ta{n}^{-1}\sqrt{3}-\pi +co{t}^{-1}\sqrt{3}& \\ =ta{n}^{-1}\sqrt{3}-\pi +ta{n}^{-1}\frac{1}{\sqrt{3}}& (\because ta{n}^{-1}\frac{1}{x}=co{t}^{-1}x)\\ =[ta{n}^{-1}\sqrt{3}+ta{n}^{-1}\frac{1}{\sqrt{3}}]-\pi & \\ =ta{n}^{-1}\frac{(\sqrt{3}+\frac{1}{\sqrt{3}})}{1-(\sqrt{3}\times \frac{1}{\sqrt{3}})}-\pi & [\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)]\\ =ta{n}^{-1}\left(\infty \right)-\pi & \\ =\frac{\pi }{2}-\pi =-\frac{\pi }{2}& (\because tan\frac{\pi }{2}=\infty )\end{array}$

Hence, the correct option is (b).