Find the value of$\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$ is equal to

Question:

Find the value of $\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$ is equal to

(A) $\pi(\mathbf

{B})-\frac{\pi}{3}$

(C) $\frac{\pi}{3}$

(D) $\frac{2 \pi}{3}$

Solution:

Let $\tan ^{-1} \sqrt{3}=x$. Then, $\tan x=\sqrt{3}=\tan \frac{\pi}{3}$.

We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$.

$\therefore \tan ^{-1} \sqrt{3}=\frac{\pi}{3}$

Let $\sec ^{-1}(-2)=y$. Then, $\sec y=-2=-\sec \left(\frac{\pi}{3}\right)=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \frac{2 \pi}{3}$.

We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$.

$\therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}$

Hence, $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}$

 

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