Question

Find the value of \(\tan^{-1}\sqrt3-\sec^{-1}(-2)\).

Hint:

Use tan^-1(root3) = pi/3, and sec^-1(-2) = pi - sec^-1(2) = 2pi/3.

Answer 1

Step 1: Evaluate tan inverse root 3:
The principal value branch of \(\tan^{-1}\) is \(\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\).
Since \(\tan\dfrac{\pi}{3} = \sqrt3\), we get \(\tan^{-1}\sqrt3 = \dfrac{\pi}{3}\).

Step 2: Evaluate sec inverse of -2:
The principal value branch of \(\sec^{-1}\) is \([0,\pi]\), excluding \(\dfrac{\pi}{2}\).
For a negative input the angle lies in \(\left(\dfrac{\pi}{2}, \pi\right)\). Since \(\sec\dfrac{2\pi}{3} = \dfrac{1}{\cos(2\pi/3)} = \dfrac{1}{-1/2} = -2\), we get:
\[ \sec^{-1}(-2) = \frac{2\pi}{3} \]

Step 3: Subtract the two values:
Substitute both principal values into the expression:
\[ \tan^{-1}\sqrt3 - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \]

Final Answer:
The value of the expression is negative pi by 3. \[ \boxed{-\frac{\pi}{3}} \]