Question

The value of $\tan^{-1}(A) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$ is}

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{2}$
• C. $\frac{3\pi}{4}$
• D. $\frac{3\pi}{2}$

Hint:

Remember:
• $\tan^{-1}(A) = \frac{\pi}{4}$
• $\cos^{-1}(-x)$ lies in second quadrant
• $\sin^{-1}(-x)$ lies in fourth quadrant

Answer 1

The Correct Option is C

Concept: Inverse trigonometric functions return principal values within restricted ranges. To evaluate a combination of inverse trigonometric expressions, we compute each term separately using standard angles and then add the results carefully.

Step 1: Evaluate $\tan^{-1(A)$}
We know that: \[ \tan\left(\frac{\pi}{4}\right) = 1 \] Therefore, \[ \tan^{-1}(A) = \frac{\pi}{4} \]

Step 2: Evaluate $\cos^{-1\left(-\frac{1}{2}\right)$}
We know: \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Since cosine is negative, the angle lies in the second quadrant: \[ \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \]

Step 3: Evaluate $\sin^{-1\left(-\frac{1}{2}\right)$}
We know: \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] Since sine is negative, principal value lies in fourth quadrant: \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \]

Step 4: Add all values
\[ E = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} \] Taking LCM = 12: \[ E = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} \] \[ E = \frac{9\pi}{12} = \frac{3\pi}{4} \] Thus, the final answer is $\frac{3\pi}{4}$.