COMEDK · Maths · 26. Differentiation
If \(\quad 2 y=\left[\cot ^{-1}\left(\dfrac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right]^2 \forall x \in\left(0, \dfrac{\pi}{2}\right)\) then \(\dfrac{d y}{d x}\) is equal to
- A \(x-\dfrac{\pi}{6}\)
- B \(\dfrac{\pi}{3}-x\)
- C \(2 x-\dfrac{\pi}{3}\)
- D \(\dfrac{\pi}{6}-x\)
Answer & Solution
Correct Answer
(A) \(x-\dfrac{\pi}{6}\)
Step-by-step Solution
Detailed explanation
The given expression is \(2y = \left[ \cot^{-1} \left( \dfrac{\sqrt{3} \cos x + \sin x}{\cos x - \sqrt{3} \sin x} \right) \right]^2\). Divide the numerator and denominator inside the \(\cot^{-1}\) function by \(\cos x\):…
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