#### Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 19 Maths Textbook Solution.

Answer: $3y=2 \sqrt{2x}-2$

Hint: Differentiate to find the slop of the tangent

Given$x=\sin \: 3t,y=\cos t2t$ at $t=\frac{\pi }{4}$

Solution: Slope of tangent is $\frac{dy}{dx}$

\begin{aligned} &\Rightarrow \frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{d(\cos 2 t)}{d t}}{\frac{d(\sin 3 t)}{d t}}=\frac{-2 \sin 2 t}{3 \cos 3 t} \\ &{\left[\frac{d y}{d x}\right]_{x t-\frac{\pi}{4}}=\frac{-2 \sin \frac{\pi}{2}}{3 \cos \frac{3 \pi}{4}}=\frac{-2 \times 1}{3 \times\left[-\frac{1}{\sqrt{2}}\right]}=\frac{2 \sqrt{2}}{3}} \end{aligned}

Now,$x=\sin \left [ \frac{3\pi }{4} \right ]=\frac{1}{\sqrt{2}}$

$y=\cos \left [ \frac{2\pi }{4} \right ]=0$

$\therefore$ Equation of the tangent is

\begin{aligned} &y-0=\frac{d y}{d x}\left[x-\left[\frac{1}{\sqrt{2}}\right]\right] \\ &y=\frac{2 \sqrt{2}}{3}\left[x-\frac{1}{\sqrt{2}}\right] \\ &y=\frac{2 \sqrt{2}}{3} x-\frac{2}{3} \end{aligned}

or $3y=2\sqrt{2x}-2$