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#### Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 23 Maths Textbook Solution.

$\frac{d y}{d x}\; \: \mathrm{At}\; \; \left(\theta=\frac{\pi}{3}\right)=-\sqrt{3}$

Hint:

Use chain rule and   $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

Given:

\begin{aligned} &x=a(\theta-\sin \theta) \\ &y=a(1+\cos \theta) \end{aligned}

Solution:

$x=a(\theta-\sin \theta) \\$

$\frac{d x}{d \theta}=a\left[\frac{d \theta}{d \theta}-\frac{d \sin \theta}{d \theta}\right] \\$                                                            $\left[\frac{d \sin \theta}{d \theta}=\cos \theta\right]$

$\frac{d x}{d \theta}=a(1-\cos \theta) \\$                                                        (1)

\begin{aligned} & &y=a(1+\cos \theta) \end{aligned}

$\frac{d y}{d \theta}=a\left(\frac{d 1}{d \theta}+\frac{d \cos \theta}{d \theta}\right) \\$

$=a(0+(-\sin \theta)) \\$                                                                                     $\left[\because \frac{d \cos \theta}{d \theta}=-\sin \theta\right]$

\begin{aligned} & &\frac{d y}{d \theta}=-a \sin \theta \end{aligned}                                                                                                               (2)

$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

Put the values of  $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$  from the equation (2) and (1) respectively

$\frac{d y}{d x}=\frac{-a \sin \theta}{a(1-\cos \theta)} \\$

$\frac{d y}{d x}=\frac{-\sin \theta}{1-\cos \theta} \\$

\begin{aligned} &\text { At } \theta=\frac{\pi}{3} \end{aligned}

$\frac{d y}{d x}=-\left(\frac{\sin \frac{\pi}{3}}{1-\cos \frac{\pi}{3}}\right)$

$=-\left(\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}\right)$                                                                       $\left[\begin{array}{rl} \because \sin \frac{\pi}{3} & =\frac{\sqrt{3}}{2} \\\\ \cos \frac{\pi}{3} & =\frac{1}{2} \end{array}\right]$

$=-\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) \\$

\begin{aligned} & &\frac{d y}{d x} a t \theta=\frac{\pi}{3}=-\sqrt{3} \end{aligned}