Question

Differentiate (1+cosx) / (1-cosx) with respect to x.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{y}=\frac{1+\cos \mathrm{x}}{1-\cos \mathrm{x}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(1-\cos \mathrm{x}\right)\times {\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(1+\cos \mathrm{x}\right)-\left(1+\cos \mathrm{x}\right)\times {\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(1-\cos \mathrm{x}\right)}{{\left(1-\cos \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(1-\cos \mathrm{x}\right)\left(-\sin \mathrm{x}\right)-\left(1+\cos \mathrm{x}\right)\left(\sin \mathrm{x}\right)}{{\left(1-\cos \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\sin \mathrm{x}+\sin \mathrm{x}.\cos \mathrm{x}-\sin \mathrm{x}-\sin \mathrm{x}.\cos \mathrm{x}}{{\left(1-\cos \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2\sin \mathrm{x}}{{\left(1-\cos \mathrm{x}\right)}^2} \end{gathered}$$