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Question Number 64514 by Mikael last updated on 18/Jul/19

$$y=arctan\left[\sqrt{\frac{1-cosx}{1+cosx}}\right]$$$$y^{}=?$$

Commented by mathmax by abdo last updated on 18/Jul/19

$$wehave\frac{1-cosx}{1+cosx}=\frac{2{sin}^2\left(\frac{x}{2}\right)}{2{cos}^2\left(\frac{x}{2}\right)}={tan}^2\left(\frac{x}{2}\right)\Rightarrow \sqrt{\frac{1-cosx}{1+cosx}}=\mid tan\left(\frac{x}{2}\right)\mid$$$$case1tan\left(\frac{x}{2}\right)>0\Rightarrow y\left(x\right)=\frac{x}{2}\Rightarrow y^{{}^{\prime }}\left(x\right)=\frac{1}{2}$$$$case2tan\left(\frac{x}{2}\right)<0\Rightarrow y\left(x\right)=-\frac{x}{2}\Rightarrow y^{{}^{\prime }}\left(x\right)=-\frac{1}{2}$$

Commented by Mikael last updated on 19/Jul/19

$$thankyouSir$$

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