#### Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 16

$\frac{d y}{d x}=\frac{1}{\sqrt{3}} \; \; \text { At }\; \; t=\frac{2 \pi}{3}$

Hint:

Use   $[\tan (\pi-\theta)=-\tan \theta]$   and   $\frac{d(\sin t)}{d t}=\cos t$

Given:

\begin{aligned} &x=\cos t \\ &y=\sin t \end{aligned}

Solution:

$x=\cos t \\$

\begin{aligned} & &\frac{d x}{d t}=\frac{d \cos t}{d t}=-\sin t \end{aligned}                                                                                                               (1)

$y=\sin t$

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

\begin{aligned} & &\frac{d y}{d t}=\cos t \end{aligned}                                                                                                                                 (2)

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

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

$=\frac{\cos t}{-\sin t}$

$\frac{d y}{d x}\; \; At\; \; \left(x=\frac{2 \pi}{3}\right)$

\begin{aligned} &=-\cot \left(\frac{2 \pi}{3}\right) \\ &=-\cot \left(\pi-\frac{\pi}{3}\right) \\ &=-\left(-\cot \frac{\pi}{3}\right) \\ &=\cot \frac{\pi}{3} \end{aligned}

$\frac{d y}{d x}\;\; At\; \; \left(x=\frac{2 \pi}{3}\right)=\frac{1}{\sqrt{3}}$

Hence proved