#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 12 maths

$\sqrt{\frac{1-y^{2}}{1-x^{2}}}$

Hint:

Use  $(\sin A-\sin B) \text { and }(\cos A+\cos B)$ to formulas and use $\frac{d\left(\sin ^{-1} x\right)}{d x}=\frac{1}{\sqrt{1-x^{2}}}$

Given:

$\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$

Solution:

Let  $x=\sin A, y=\sin B$

So, the equation will become

$\sqrt{1-\sin ^{2} A}+\sqrt{1-\sin ^{2} B}=a(\sin A-\sin B)$

$\sqrt{\cos ^{2} A}+\sqrt{\cos ^{2} B}=a(\sin A-\sin B) \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

\begin{aligned} &\cos A+\cos B=a(\sin A-\sin B) \\ &a=\frac{\cos A+\cos B}{\sin A-\sin B} \end{aligned}

$a=\frac{2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)}$                    $\left[\begin{array}{l} \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right) \\\\ \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right) \end{array}\right]$

$a=\cot \left(\frac{A-B}{2}\right)$

\begin{aligned} &\cot ^{-1} a=\frac{A-B}{2} \\ &2 \cot ^{-1} a=A-B \end{aligned}                                    $\left[\because \frac{\cos \theta}{\sin \theta}=\cot \theta\right]$

$2 \cot ^{-1} a=\sin ^{-1} x-\sin ^{-1} y \quad[x=\sin A, y=\sin B]$

Differentiate the above equation w.r.t x

$\frac{d\left(2 \cot ^{-1} a\right)}{d x}=\frac{d\left(\sin ^{-1} x-\sin ^{-1} y\right)}{d x}$

$2 \frac{d\left(\cot ^{-1} a\right)}{d x}=\frac{d\left(\sin ^{-1} x\right)}{d x}-\frac{d\left(\sin ^{-1} y\right)}{d x}$

$2 \times 0=\frac{1}{\sqrt{1-x^{2}}}-\frac{d\left(\sin ^{-1} y\right)}{d y} \times \frac{d y}{d x}$            $\left[\because \frac{d\left(\sin ^{-1} x\right)}{d x}=\frac{1}{\sqrt{1-x^{2}}}, \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$0=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}$

$\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}$

$\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$

Hence, if $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$

Then,$\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$ is the required answer

Hence proved