Question

proof that,..

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plz gave me right answer ..​

Answer 1

Step-by-step explanation:

$2 tan^{-1}(\frac{1 + x}{1 - x}) + sin^{-1}(\frac{1 - x^2}{1 + x^2})$

$=> 2 tan^{-1}(\frac{1 + tan\theta}{1 - tan\theta}) + sin^{-1}(\frac{1 - (tan\theta)^2}{1 + (tan\theta)^2})$

$=> 2 tan^{-1}(\frac{tan(\frac{\pi }{4}) + tan\theta}{1 - tan(\frac{\pi }{4})tan\theta}) + sin^{-1}(\frac{1 - (tan\theta)^2}{1 + (tan\theta)^2})$

$=> 2 tan^{-1} (tan(\frac{\pi }{4} + \theta)) + sin^{-1}(cos 2\theta)}$

$=> 2 (\frac{\pi }{4} + \theta) + sin^{-1}(sin(90 -  2\theta)})$

$=> \frac{\pi }{2} + 2\theta + 90 - 2\theta$

$=> \frac{\pi }{2} + 90$

$=> \frac{\pi }{2} + \frac{\pi }{2}$

$=> \frac{2\pi }{2}$

$=> \pi$

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