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Question Number 41536 by mondodotto@gmail.com last updated on 09/Aug/18

$$\mathbf{if}\mathit{y}=\sqrt{\frac{1+\mathbf{sin}\mathit{x}}{1-\mathbf{sin}\mathit{x}}}\mathbf{show}\mathbf{that}$$$$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{1}{1-\mathbf{sin}\mathit{x}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

$$sinx=\frac{2tan\frac{x}{2}}{1+{tan}^2\frac{x}{2}}$$$$y=\sqrt{\frac{1+\frac{2tan\frac{x}{2}}{1+{tan}^2\frac{x}{2}}}{1-\frac{2tan\frac{x}{2}}{1+{tan}^2\frac{x}{2}}}}$$$$y=\sqrt{\frac{{(1+tan\frac{x}{2})}^2}{{(1-tan\frac{x}{2})}^2}}$$$$y=\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}=tan(\frac{\mathrm{\Pi }}{4}+\frac{x}{2})$$$$\frac{dy}{dx}={sec}^2(\frac{\mathrm{\Pi }}{4}+\frac{x}{2}).\frac{1}{2}$$$$\frac{dy}{dx}=\frac{1}{2{cos}^2(\frac{\mathrm{\Pi }}{4}+\frac{x}{2})}=\frac{1}{1+cos\{2.(\frac{\mathrm{\Pi }}{4}+\frac{x}{2})\}}=\frac{1}{1+cos(\frac{\mathrm{\Pi }}{2}+x)}$$$$\frac{dy}{dx}=\frac{1}{1-sinx}proved$$

Answered by math1967 last updated on 09/Aug/18

$$y=\sqrt{\frac{(1+sinx)(1-sinx)}{{(1-sinx)}^2}}$$$$y=\frac{cosx}{1-sinx}$$$$\frac{dy}{dx}=\frac{-sinx(1-sinx)-cosx.(-cosx)}{{(1-sinx)}^2}$$$$\frac{dy}{dx}=\frac{-sinx+{sin}^{2}x+{cos}^{2}x}{{(1-sinx)}^2}=\frac{1-sinx}{{(1-sinx)}^2}$$$$\frac{dy}{dx}=\frac{1}{1-sinx}$$$$$$

Answered by $@ty@m last updated on 09/Aug/18

$$Itcanbeshownthat$$$$y=\frac{1+\sin x}{\cos x}$$$$y=\sec x+\tan x$$$$\frac{dy}{dx}=\sec x\tan x+\sec {}^{2}x$$$$=\sec x(\sec x+\tan x)$$$$=\frac{1+\sin x}{\cos {}^{2}x}$$$$=\frac{1}{1-\sin x}$$

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