Question

The value of \[{{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)\] is equal to:
(a) $\pi $
(b) $-\dfrac{\pi }{3}$
(c) $\dfrac{\pi }{3}$
(d) $\dfrac{2\pi }{3}$

Answer 1

Hint: Find the principal value of \[{{\tan }^{-1}}\sqrt{3}\] and \[{{\sec }^{-1}}\left( -2 \right)\] by considering the proper range of the given inverse functions and take their difference to get the answer. Use the formula: \[{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x\] and then find the principal value of \[{{\sec }^{-1}}\left( 2 \right)\].

Complete step by step answer:
We have been given to find the value of the expression: \[{{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)\].
Let us find the principal value of \[{{\tan }^{-1}}\sqrt{3}\].
We know that the range of \[{{\tan }^{-1}}x\] is between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. So, we have to consider such an angle whose tangent is $\sqrt{3}$ and it lies in the range $-\dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$.
We know that, $\tan \dfrac{\pi }{3}=\sqrt{3}$, here $\dfrac{\pi }{3}$ is in the range $-\dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$.
Therefore, the principal value of \[{{\tan }^{-1}}\sqrt{3}\] is $\dfrac{\pi }{3}$.
Now, let us find the principal value of \[{{\sec }^{-1}}\left( -2 \right)\].
Using the formula, \[{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x\], we get,
\[{{\sec }^{-1}}\left( -2 \right)=\pi -{{\sec }^{-1}}2\]
So, we have to find the principal value of \[{{\sec }^{-1}}2\].
We know that range of \[{{\sec }^{-1}}x\] is between \[0\] to $\pi $, excluding $\dfrac{\pi }{2}$. So, we have to consider such an angle whose secant is 2 and it lies in the range of \[0\] to $\pi $.
We know that, $\sec \dfrac{\pi }{3}=2$, here, $\dfrac{\pi }{3}$ is in the range \[0\] to $\pi $.
Therefore, the principal value of \[{{\sec }^{-1}}2\] is $\dfrac{\pi }{3}$.
Now, the expression
\[\begin{align}
  & {{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right) \\
 & =\dfrac{\pi }{3}-\left( \pi -\dfrac{\pi }{3} \right) \\
 & =-\pi +\dfrac{\pi }{3}+\dfrac{\pi }{3} \\
 & =-\pi +\dfrac{2\pi }{3} \\
 & =\dfrac{-3\pi +2\pi }{3} \\
 & =\dfrac{-\pi }{3} \\
\end{align}\]
Hence, option (b) is the correct answer.

Note: One may note that there is only one principal value of an inverse trigonometric function. We know that at many angles the value of tan and sec are $\sqrt{3}$ and -2 respectively, but we have to remember the range in which these inverse functions are defined. We have to choose such an angle which lies in the range and satisfies the function. So, there can be only one answer.