#### Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 22 Maths Textbook Solution.

$\frac{d y}{d x}=\frac{6}{5} \cot \left(\frac{t}{2}\right)$

Hint:

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

Given:

\begin{aligned} &x=10(t-\sin t) \\ &y=12(1-\cos t) \end{aligned}

Solution:

$x=10(t-\sin t) \\$

\begin{aligned} & &\frac{d x}{d t}=10\left(\frac{d t}{d t}-\frac{d \sin t}{d t}\right) \end{aligned}                                     $\left[\frac{d \sin t}{d t}=\cos t\right]$

$\frac{d x}{d t}=10(1-\cos t) \\$                                                                                                       (1)

$y=12(1-\cos t) \\$

$\frac{d y}{d t}=12\left(\frac{d(1)}{d t}-\frac{d \cos t}{d t}\right) \\$

\begin{aligned} & &=12(0-(-\sin t)) \end{aligned}                                           $\left[\because \frac{d \cos t}{t}=-\sin t\right]$

$\frac{d y}{d t}=12 \sin t \\$                                                                                                                (2)

\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}

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

$\frac{d y}{d x}=\frac{12 \sin t}{10(1-\cos t)}=\frac{6 \sin t}{5(1-\cos t)} \\$

\begin{aligned} & &=\frac{6}{5} \times \frac{2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \sin ^{2} \frac{t}{2}} \end{aligned}

$=\frac{6}{5} \times \frac{2 \times \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \times \sin \frac{t}{2} \cdot \sin \frac{t}{2}} \\$

\begin{aligned} & &=\frac{6}{5} \cot \frac{t}{2} \end{aligned}                                                                                                                            $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$