Question

If ${\tan }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=a$, prove that $\frac{dy}{dx}=\frac{x}{y}\frac{\left(1-\tan a\right)}{\left(1+\tan a\right)}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},{\tan }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=a \\ \Rightarrow \frac{x^2-y^2}{x^2+y^2}=\tan a \\ \Rightarrow x^2-y^2=\tan a\left(x^2+y^2\right) \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(x^2-y^2\right)=\tan a\frac{d}{dx}\left(x^2+y^2\right) \\ \Rightarrow \left(2x-2y\frac{\mathit{d}y}{\mathit{d}x}\right)=\tan a\left(2x+2y\frac{\mathit{d}y}{\mathit{d}x}\right) \\ \Rightarrow 2x-2y\frac{\mathit{d}y}{\mathit{d}x}=2x\tan a+2y\tan a\frac{\mathit{d}y}{\mathit{d}x} \\ \Rightarrow 2y\tan a\frac{\mathit{d}y}{\mathit{d}x}+2y\frac{\mathit{d}y}{\mathit{d}x}=2x-2x\tan a \\ \Rightarrow 2y\left(1+\tan a\right)\frac{\mathit{d}y}{\mathit{d}x}=2x\left(1-\tan a\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{x}{y}\left(\frac{1-\tan a}{1+\tan a}\right) \end{gathered}$$
Hence proved