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Question
If \[\tan^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = a\] Prove that \[\frac{dy}{dx} = \frac{x}{y}\frac{\left( 1 - \tan a \right)}{\left( 1 + \tan a \right)}\] ?
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Solution
\[\text{ We have }, \tan^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = a\]
\[ \Rightarrow \frac{x^2 - y^2}{x^2 + y^2} = \tan a\]
\[ \Rightarrow x^2 - y^2 = \tan a\left( x^2 + y^2 \right)\]
Differentiating with respect to x,
\[\Rightarrow \frac{d}{dx}\left( x^2 - y^2 \right) = \tan a\frac{d}{dx}\left( x^2 + y^2 \right)\]
\[ \Rightarrow \left( 2x - 2y\frac{d y}{d x} \right) = \tan a\left( 2x + 2y\frac{d y}{d x} \right)\]
\[ \Rightarrow 2x - 2y\frac{d y}{d x} = 2x\tan a + 2y\tan a\frac{d y}{d x}\]
\[ \Rightarrow 2y\tan a\frac{d y}{d x} + 2y\frac{d y}{d x} = 2x - 2x\tan a\]
\[ \Rightarrow 2y\left( 1 + \tan a \right)\frac{d y}{d x} = 2x\left( 1 - \tan a \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{x}{y}\left( \frac{1 - \tan a}{1 + \tan a} \right)\]
Hence proved
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