Question

The value of \(\frac{tan^{-1}(1)+cos^{-1}(\frac{1}{2})+sin^{-1}(\frac{1}{2})}{tan^{-1}(\sqrt{3})-sec^{-1}(-2)}\) is equal to

• A. \(\frac{3}{4}\)
• B. \(\frac{-1}{3}\)
• C. \(-4\)
• D. \(\frac{-9}{4}\)

Hint:

Evaluate each principal value in terms of pi, remembering \(\sec^{-1}(-2) = \pi - \sec^{-1}2\).

Answer 1

The Correct Option is D

Step 1: Numerator:
\(\tan^{-1}1 = \frac{\pi}{4}\), \(\cos^{-1}\frac12 = \frac{\pi}{3}\), \(\sin^{-1}\frac12 = \frac{\pi}{6}\).
Sum \(= \frac{\pi}{4}+\frac{\pi}{3}+\frac{\pi}{6} = \frac{3\pi+4\pi+2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}\).

Step 2: Denominator:
\(\tan^{-1}\sqrt3 = \frac{\pi}{3}\). Also \(\sec^{-1}(-2) = \pi - \sec^{-1}2 = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\), since \(\sec^{-1}\) of a negative number lies in the second quadrant.
Denominator \(= \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}\).

Step 3: Divide:
\[ \frac{3\pi/4}{-\pi/3} = -\frac{9}{4} \]

Final Answer:
The value is \(-\frac{9}{4}\), option (D). \[ \boxed{-\frac{9}{4}} \]