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Question Number 30528 by abdo imad last updated on 22/Feb/18

$$simplifyA=arctan\left(\frac{sinx}{1-cosx}\right).$$

Commented by abdo imad last updated on 23/Feb/18

$$wehaveA=arctan\left(\frac{2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}{2{sin}^2\left(\frac{x}{2}\right)}\right)$$$$A=arctan\left(cotan\left(\frac{x}{2}\right)\right)=arctan\left(\frac{1}{tan\left(\frac{x}{2}\right)}\right)$$$$case1if0<x<\pi tan\left(\frac{x}{2}\right)>0\Rightarrow A=\frac{\pi }{2}-arctan\left(tan\left(\frac{x}{2}\right)\right)$$$$\Rightarrow A=\frac{\pi }{2}-\frac{x}{2}=\frac{\pi -x}{2}$$$$case2if-\pi <x<0tan\left(\frac{x}{2}\right)<0\Rightarrow A=-\frac{\pi }{2}-arctan\left(tan\left(\frac{x}{2}\right)\right)$$$$\Rightarrow A=-\frac{\pi }{2}-\frac{x}{2}.$$

Answered by prakash jain last updated on 22/Feb/18

$$\frac{\sin x}{1-\cos x}=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{{2\sin }^2\frac{x}{2}}=\cot \frac{x}{2}$$$$arctan\left(\frac{sinx}{1-cosx}\right)={\tan }^{-1}\cot \frac{x}{2}$$

Commented by prakash jain last updated on 23/Feb/18

$$\mathrm{what}\mathrm{if}x=8\pi ,9\pi \mathrm{etc}?$$$${\tan }^{-1}x\mathrm{should}\mathrm{be}\mathrm{from}-\frac{\pi }{2}\mathrm{to}\frac{\pi }{2}$$

Commented by mrW2 last updated on 23/Feb/18

$${\tan }^{-1}\cot \frac{x}{2}=\frac{\pi }{2}-\frac{x}{2}???$$

Commented by mrW2 last updated on 23/Feb/18

$$for0<x<2\pi :$$$$A=\frac{1}{2}(\pi -x)$$$$$$$$generally:$$$$A=\frac{1}{2}(\pi +2\pi \left[\frac{x}{2\pi }\right]-x)$$$$$$$$e.g.:$$$$x=6.5\pi$$$$\left[\frac{x}{2\pi }\right]=\left[\frac{6.5\pi }{2\pi }\right]=3$$$$A=\frac{1}{2}(\pi +6\pi -6.5\pi )=\frac{\pi }{4}$$

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