Question

Find the principal value of ${\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2).$

Answer 1

Let ${\tan }^{-1}\sqrt{3}=x$ that is

$\tan x=\sqrt{3}=\tan \frac{\pi }{3}$

Therefore, $x=\frac{\pi }{3}\in [-\frac{\pi }{2},\frac{\pi }{2}]$

Now let ${\sec }^{-1}(-2)=y$ that is $\sec y=-2$.

Therefore, $y=-\sec \frac{\pi }{3}$ that is

$\sec y=\sec (\pi -\frac{\pi }{3})=\sec \frac{2\pi }{3}$

Thus, $y=\frac{2\pi }{3}\in [0,\pi ]-\left(\frac{\pi }{2}\right)$

Now consider ${\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)$ as shown below:

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

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