Question

Differentiate

$\log \sqrt{\frac{1-\cos x}{1+\cos x}}$

Answer 1

$$\begin{gathered} \mathrm{Let}y=\log \sqrt{\frac{1-\cos x}{1+\cos x}} \\ \Rightarrow y=\log {\left(\frac{1-\cos x}{1+\cos x}\right)}^{\frac{1}{2}} \\ \Rightarrow y=\frac{1}{2}\log \left(\frac{1-\cos x}{1+\cos x}\right)\left[\mathrm{using}\log a^b=b\log a\right] \\ \mathrm{Differentiate}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x\mathrm{we}\mathrm{get}, \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}\left\{\frac{1}{2}\log \left(\frac{1-\cos x}{1+\cos x}\right)\right\} \\ =\frac{1}{2}\times \frac{1}{\left({\displaystyle \frac{1-\cos x}{1+\cos x}}\right)}\times \frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{1}{2}\left(\frac{1+\cos x}{1-\cos x}\right)\left[\frac{\left(1+\cos x\right){\displaystyle \frac{d}{dx}}\left(1-\cos x\right)-\left(1-\cos x\right){\displaystyle \frac{d}{dx}}\left(1+\cos x\right)}{{\left(1+\cos x\right)}^2}\right]\left[\mathrm{Using}\mathrm{quotient}\mathrm{rule}\right] \\ =\frac{1}{2}\left(\frac{1+\cos x}{1-\cos x}\right)\left[\frac{\left(1+\cos x\right)\left(\sin x\right)-\left(1-\cos x\right)\left(-\sin x\right)}{{\left(1+\cos x\right)}^2}\right] \\ =\frac{1}{2}\left(\frac{1+\cos x}{1-\cos x}\right)\left[\frac{\sin x+\sin x\cos x+\sin x-\sin x\cos x}{{\left(1+\cos x\right)}^2}\right] \\ =\frac{1}{2}\left(\frac{1+\cos x}{1-\cos x}\right)\left[\frac{2\sin x}{{\left(1+\cos x\right)}^2}\right] \\ =\frac{\sin x}{\left(1-\cos x\right)\left(1+\cos x\right)} \\ =\frac{\sin x}{1-{\cos }^{2}x} \\ =\frac{\sin x}{{\sin }^{2}x} \\ =\frac{1}{\sin x} \\ =\mathrm{cosec}x \\ \mathrm{So},\frac{d}{dx}\left(\log \sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\mathrm{cosec}x \end{gathered}$$