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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 $\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{1-\text{x}^2}$
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Answer

We have, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}\big(\text{x}-\text{y}\big)$
Let $\text{x}=\sin\text{A},\text{y}=\sin\text{B}$
$\Rightarrow\sqrt{1-\sin^2\text{A}}+\sqrt{1-\sin^2\text{B}}=\text{a}\big(\sin\text{A}-\sin\text{B}\big)$
$\Rightarrow\cos\text{A}+\cos\text{B}=\text{a}\big(\sin\text{A}-\sin\text{B}\big)$
$\Rightarrow\text{a}=\frac{\cos\text{A}+\cos\text{B}}{\sin\text{A}-\sin\text{B}}$
$\Rightarrow\text{a}=\frac{2\cos\frac{\text{A}+\cos\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}}{2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2}}$
$\begin{bmatrix}\because\sin\text{A}-\sin\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2} \\ \because\cos \text{A}+\cos\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}\end{bmatrix}$
$\Rightarrow\text{a}=\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$\Rightarrow\cot^{-1}\text{a}=\frac{\text{A}-\text{B}}{2}$
$\Rightarrow2\cot^{-1}\text{a}={\text{A}-\text{B}}$
$\Rightarrow2\cot^{-1}\text{a}=\sin^{-1}\text{x}-\sin^{-1}\text{y}$
$\Big[\because \text{x}=\sin\text{A},\text{y}=\sin\text{B}\Big]$
Differentiating with respect to x, we get
$\frac{\text{d}}{\text{dx}}\big(2\cot^{-1}\text{a}\big)=\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)-\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{y}\big)$
$\Rightarrow0=\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sqrt\frac{1-\text{y}^2}{1-\text{x}^2}$

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