Question

The value of \(sin^{-1}(sin\frac{5π}{6})+cos^{-1}(cos\frac{7π}{6})+tan^{-1}(tan\frac{2π}{3})\) is equal to...

• A. \(\frac{2π}{3}\)
• B. \(\frac{4π}{3}\)
• C. \(\frac{5π}{3}\)
• D. \(π\)

Hint:

Bring each angle into the principal range before applying the inverse function.

Answer 1

The Correct Option is A

Step 1: Understand the concept
The principal ranges are \(\sin^{-1}\): \([-\frac{\pi}{2}, \frac{\pi}{2}]\), \(\cos^{-1}\): \([0, \pi]\) and \(\tan^{-1}\): \((-\frac{\pi}{2}, \frac{\pi}{2})\). We must shift each angle into the right range.

Step 2: First term
\(\sin\frac{5\pi}{6} = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\frac{\pi}{6}\), so \(\sin^{-1}(\ldots) = \frac{\pi}{6}\).

Step 3: Second and third terms
\(\cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{5\pi}{6}\right) = \cos\frac{5\pi}{6}\), and \(\frac{5\pi}{6}\) is in \([0, \pi]\), so it gives \(\frac{5\pi}{6}\). \(\tan\frac{2\pi}{3} = \tan\left(\frac{2\pi}{3} - \pi\right) = \tan\left(-\frac{\pi}{3}\right)\), giving \(-\frac{\pi}{3}\).

Step 4: Add
\[ \frac{\pi}{6} + \frac{5\pi}{6} - \frac{\pi}{3} = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
Naively cancelling the inverse and the function would give \(\frac{5\pi}{6} + \frac{7\pi}{6} + \frac{2\pi}{3} = \frac{8\pi}{3}\), which is wrong because the angles are outside the principal ranges.

Final Answer:
The sum is 2 pi / 3. This is option (A). \[ \boxed{\text{(A) }\frac{2\pi}{3}} \]