Question

The value of the expression:
${\tan }^{-1}\sqrt{3}-{\cot }^{-1}(-\sqrt{3})$ is equal to

• A. $0$
• B. $-\frac{\pi }{2}$
• C. $2\sqrt{3}$
• D. $\pi$

Answer 1

The correct option is B $-\frac{\pi }{2}$

Given :
${\tan }^{-1}\sqrt{3}-{\cot }^{-1}(-\sqrt{3})$
Let $y={\tan }^{-1}\left(\sqrt{3}\right)$
$\Rightarrow \tan y=\sqrt{3}$
$\Rightarrow \tan y=\tan \left(\frac{\pi }{3}\right)$
We know that the range of the principal value branch of ${\tan }^{-1}$ is $(\frac{-\pi }{2},\frac{\pi }{2})$.
$\therefore y=\frac{\pi }{3}\Rightarrow {\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$
Let $z={\cot }^{-1}(-\sqrt{3})$
$\Rightarrow \cot z=-\sqrt{3}$
$\Rightarrow \cot z=\cot \left(\frac{-\pi }{6}\right)$
We know that the range of the principal value branch of ${\cot }^{-1}$ is $(0,\pi )$.
$\therefore z=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$
$\Rightarrow {\cot }^{-1}(-\sqrt{3})=\frac{5\pi }{6}$
Now ${\tan }^{-1}\sqrt{3}-{\cot }^{-1}(-\sqrt{3})$
$=\frac{\pi }{3}-\frac{5\pi }{6}=-\frac{\pi }{2}$
Hence, the correct option is $\left(B\right).$