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Question: If \(\tan^{- 1}\frac{\sqrt{1 - x^{2}} - 1}{x} = 4\), then...

If tan⁡−11−x2−1x=4\tan^{- 1}\frac{\sqrt{1 - x^{2}} - 1}{x} = 4, then

A

x=tan⁡2x = \tan 2

B

x=tan⁡4x = \tan 4

C

x = tan(1/4)

D

x=tan⁡8x = \tan 8

Answer

x=tan⁡8x = \tan 8

Explanation

Solution

Taking x=tan⁡θ,tan⁡−11−x2−1x=tan⁡−1sec⁡θ−1tan⁡θx = \tan\theta,\tan^{- 1}\frac{\sqrt{1 - x^{2}} - 1}{x} = \tan^{- 1}\frac{\sec\theta - 1}{\tan\theta}

= tan⁡−11−cos⁡θsin⁡θ=tan⁡−1(tan⁡θ2)=(12)θ=(12)tan⁡−1x\tan^{- 1}\frac{1 - \cos\theta}{\sin\theta} = \tan^{- 1}\left( \tan\frac{\theta}{2} \right) = \left( \frac{1}{2} \right)\theta = \left( \frac{1}{2} \right)\tan^{- 1}x

So that according to the given condition

(12)tan⁡−1x=4\left( \frac{1}{2} \right)\tan^{- 1}x = 4 ⇒ tan⁡−1x=8\tan^{- 1}x = 8 or x=tan⁡8x = \tan 8