Question

The value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$ will be:

• A. $\pi$
• B. $-\dfrac{\pi}{3}$
• C. $\dfrac{\pi}{3}$
• D. $\dfrac{2\pi}{3}$

Hint:

For inverse trigonometric functions, remember that the range for $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and for $\sec^{-1}(x)$ is $[0, \pi] \setminus \dfrac{\pi}{2}$.

Answer 1

The Correct Option is D

We need to evaluate $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$. Let's do this step by step.

Step 1: Evaluate $\tan^{-1}(\sqrt{3})$.
The value of $\tan^{-1}(\sqrt{3})$ is the angle whose tangent is $\sqrt{3}$. We know that: \[ \tan\left(\dfrac{\pi}{3}\right) = \sqrt{3} \] So, \[ \tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}. \]

Step 2: Evaluate $\sec^{-1}(-2)$.
The value of $\sec^{-1}(-2)$ is the angle whose secant is $-2$. The secant function is the reciprocal of the cosine function, so we are looking for the angle whose cosine is $-1/2$. This corresponds to the angle: \[ \sec^{-1}(-2) = \dfrac{2\pi}{3}. \]

Step 3: Subtract the two values.
Now, we subtract the two values we calculated: \[ \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}. \]

Step 4: Conclusion.
The correct answer is (D) $\dfrac{2\pi}{3}$.