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Question Number 208624 by vipin last updated on 19/Jun/24

Answered by Berbere last updated on 19/Jun/24

$$\frac{1}{{cosec}^{-}(-\sqrt{2})}={\sin }^{-1}(-\frac{1}{\sqrt{2}})=-\frac{\pi }{4}$$$$f\left(x\right).{\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)={\tan }^{-1}\left(\frac{1-\sqrt{\frac{1-x}{1+x}}}{1+\sqrt{\frac{1-x}{1+x}}}\right)...E$$$$g\left(y\right)={\tan }^{-1}\left(\frac{1-y}{1+y}\right)=\frac{\pi }{4}-{\tan }^{-1}\left(y\right)...$$$$prof{g}^{\prime }\left(y\right)=\frac{-2}{{(1+y)}^2}.\frac{1}{1+{\left(\frac{1-y}{1+y}\right)}^2}=-\frac{1}{1+y^2}$$$$g\left(y\right)=-{\tan }^{-1}\left(y\right)+\frac{\pi }{4}$$$$f\left(x\right)=-{\tan }^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)+\frac{\pi }{4}$$$$x=cos\left(a\right)\Rightarrow \sqrt{\frac{1-x}{1+x}}=\mid \tan \left(\frac{a}{2}\right)\mid ;a\in [0,\pi ]$$$$f\left(x\right)=\frac{\pi }{4}-\frac{a}{2}=\frac{\pi }{4}-\frac{{\cos }^{-1}\left(x\right)}{2}$$$$$$$$$$$$$$

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