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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}\frac{(1-\tan\text{a})}{(1+\tan\text{a})}$
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Answer

We have, $\tan^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\tan\text{a}$
$\Rightarrow\text{x}^2-\text{y}^2=\tan\text{a}(\text{x}^2+\text{y}^2)$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{y}^2)=\tan\text{a}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)$
$\Rightarrow\Big(2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\tan\text{a}\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}\tan\text{a}+2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow2\text{y}\tan\text{a}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{x}-2\text{x}\tan\text{a}$
$\Rightarrow2\text{y}(1+\tan\text{a})\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\tan\text{a})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\Big(\frac{1-\tan\text{a}}{1+\tan\text{a}}\Big)$

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