Question:

If $y = \tan^{-1} \frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}} , $ then $ \frac{dy}{dx} $ is equal to

Updated On: Jul 6, 2022
  • $\frac{1}{1 -x^2}$
  • $\frac{1}{\sqrt{1 -x^2}}$
  • $\frac{1}{1 + x^2}$
  • $\frac{1}{\sqrt{1 + x^2}}$
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The Correct Option is B

Solution and Explanation

Put $x = \cos\theta$ $ y = \tan^{-1} \frac{\cos\theta - \sin \theta}{\cos\theta + \sin\theta} =\tan^{-1} \frac{1-\tan\theta}{1+\tan\theta} $ $=\tan^{-1} \tan\left(\frac{\pi}{4} - \theta\right) $ $= \frac{\pi}{4} - \theta = \frac{\pi}{4} -\cos^{-1} x $ $\therefore \frac{dy}{dx} = 0 - \left(\frac{1}{\sqrt{1-x^{2}}} \right) = \frac{1}{\sqrt{1-x^{2}}} $
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