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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

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

We have, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y})$
On putting $\text{x}=\sin\alpha$ and $\text{y}=\sin\beta,$ we get
$\sqrt{1-\sin^2\alpha}+\sqrt{1-\sin\beta}=\text{a}(\sin\alpha-\sin\beta)$
$\Rightarrow\ \cos\alpha+\cos\beta=\text{a}(\sin\alpha-\sin\beta)$
$\Rightarrow\ 2\cos\frac{\alpha+\beta}{2}\cdot\cos\frac{\alpha-\beta}{2}=2\text{a}\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$
$\Rightarrow\ \cos\frac{\alpha-\beta}{2}=\text{a}\sin\frac{\alpha-\beta}{2}$
$\Rightarrow\ \cot\frac{\alpha-\beta}{2}=\text{a}$
$\Rightarrow\ \frac{\alpha-\beta}{2}=\cot^{-1}\text{a}$
$\Rightarrow\ \alpha-\beta=2\cot^{-1}\text{a}$
$\Rightarrow\ \sin^{-1}\text{x}-\sin^{-1}\text{y}=2\cot^{-1}\text{a}$ $[\because\text{x}=\sin\alpha\text{ and y}=\sin\beta]$
Differentiating both sides w.r.t. x, we get
$\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\sqrt{1-\text{x}^2}}$

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