Question

.If y = 1−tanx /1+tanx, show that dy/ dx = −2 /1+sin2x

Answer 1

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