Given that (1+1+x)tan⁡y=1+1−x. Then sin 4y is equal to

tan⁡y=1+1−x1+1+x.Let x=cos⁡θ⇒tan⁡y=212+sin⁡θ2212+cos⁡θ2=sin⁡π4+sin⁡θ2cos⁡π4+cos⁡θ2=tan⁡π8+θ4⇒4y=π2+θ

Given that (1+1+x)tan⁡y=1+1−x. Then sin 4y is equal to

tan⁡y=1+1−x1+1+x.Let x=cos⁡θ⇒tan⁡y=212+sin⁡θ2212+cos⁡θ2=sin⁡π4+sin⁡θ2cos⁡π4+cos⁡θ2=tan⁡π8+θ4⇒4y=π2+θ

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Given that $(1 + \sqrt{1 + x}) \tan y = 1 + \sqrt{1 - x}$ . Then sin 4y is equal to

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Detailed Solution

$\begin{array}{l} \tan y = \frac{1 + \sqrt{1 - x}}{1 + \sqrt{1 + x}} . Let x = \cos θ \\ \Rightarrow \tan y = \frac{\sqrt{2} [\frac{1}{\sqrt{2}} + \sin \frac{θ}{2}]}{\sqrt{2} [\frac{1}{\sqrt{2}} + \cos \frac{θ}{2}]} = \frac{\sin \frac{π}{4} + \sin \frac{θ}{2}}{\cos \frac{π}{4} + \cos \frac{θ}{2}} \\ = \tan (\frac{π}{8} + \frac{θ}{4}) \Rightarrow 4 y = \frac{π}{2} + θ \end{array}$