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Question: If y = tan<sup>–1</sup>\(\frac{x - \sqrt{1 - x^{2}}}{x + \sqrt{1 - x^{2}}}\), then \(\frac{dy}{dx}\)...

If y = tan–1x−1−x2x+1−x2\frac{x - \sqrt{1 - x^{2}}}{x + \sqrt{1 - x^{2}}}, then dydx\frac{dy}{dx} is equal to –

A

−11−x2\frac{- 1}{\sqrt{1 - x^{2}}}

B

11−x2\frac{1}{\sqrt{1 - x^{2}}}

C

−x1−x2\frac{- x}{\sqrt{1 - x^{2}}}

D

None of these

Answer

11−x2\frac{1}{\sqrt{1 - x^{2}}}

Explanation

Solution

y = tan–1x−1−x2x+1−x2\frac{x - \sqrt{1 - x^{2}}}{x + \sqrt{1 - x^{2}}} let x = cos θ ⇒ θ = cos–1x

= tan–1(cos⁡θ−sin⁡θcos⁡θ+sin⁡θ)\left( \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \right)= tan–1(1−tan⁡θ1+tan⁡θ)\left( \frac{1 - \tan\theta}{1 + \tan\theta} \right)

y = tan–1(tan⁡(π4−θ))\left( \tan\left( \frac{\pi}{4} - \theta \right) \right) = π4\frac{\pi}{4} – cot–1x