Question

If \(2y = \left(\cot^{-1}\left(\frac{\sqrt{3}\cos x + \sin x}{\cos x - \sqrt{3}\sin x}\right)\right)^2,\ x \in \left(0, \frac{\pi}{2}\right)\), then \(\frac{dy}{dx}\) is equal to :

• A. \(2x - \frac{\pi}{3}\)
• B. \(\frac{\pi}{6} - x\)
• C. \(\frac{\pi}{3} - x\)
• D. \(x - \frac{\pi}{6}\)

Hint:

Divide numerator and denominator by \(\cos x\) to get \(\tan\left(x + \frac{\pi}{3}\right)\), so \(\cot^{-1}(\cdot) = \frac{\pi}{6} - x\).

Answer 1

The Correct Option is D

Divide numerator and denominator by \(\cos x\): \(\frac{\sqrt{3} + \tan x}{1 - \sqrt{3}\tan x} = \tan\left(\frac{\pi}{3} + x\right)\).
\(\cot^{-1}\left(\tan\theta\right) = \frac{\pi}{2} - \theta\), so the inner function equals \(\frac{\pi}{2} - \frac{\pi}{3} - x = \frac{\pi}{6} - x\).
Therefore \(2y = \left(\frac{\pi}{6} - x\right)^2\), i.e. \(y = \frac{1}{2}\left(\frac{\pi}{6} - x\right)^2\).
\(\frac{dy}{dx} = \left(\frac{\pi}{6} - x\right)(-1) = x - \frac{\pi}{6}\).
Answer: (4) \(x - \frac{\pi}{6}\)