#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 4 Maths Textbook Solution.

ANSWER: $(\sqrt{2}-1) x+y+\left(\frac{\pi}{4}\right)(1-\sqrt{2})=0$

HINTS:

To find equation, first find slope and one point on the tangent.

GIVEN:

$x=\theta+\sin \theta, y=1+\cos \theta \text { at } \theta=\frac{\pi}{4}$

SOLUTION:

\begin{aligned} &\because \text { Point on the tangent at } \theta=\frac{\pi}{4}\\ &\Rightarrow x=\frac{\pi}{4}+\sin \left(\frac{\pi}{4}\right)=\frac{\pi}{4}+\frac{1}{\sqrt{2}} \text { and y }=1+\cos \frac{\pi}{4}=1+\frac{1}{\sqrt{2}}\\ &\text { point } P\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}, 1+\frac{1}{\sqrt{2}}\right) \text { lies on the line } \end{aligned}

To find two slope, we find $\frac{d y}{d x}$

$\frac{d x}{d \theta}=1+\cos \theta \quad \text { and } \quad \frac{d y}{d \theta}=-\sin \theta$

Thus

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

Solving above

$\frac{d y}{d x}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \text { or } \frac{d y}{d x}=-\tan \frac{\theta}{2}$

Thus

$\frac{d y}{d x}=-\tan \left(\frac{\theta}{2}\right) \text { at } \theta=\frac{\pi}{4}$

Slope,

$\Rightarrow \mathrm{m}=-\tan \left(\frac{\pi}{8}\right)-(1-\sqrt{2)}$

Now apply point slope form to get equation of tangent,

$[y-(1+\frac{1}{\sqrt{2}})]= (1-\sqrt{2})\left(x - \frac{\pi}{4}-\frac{1}{\sqrt{2}}\right)$

$\Rightarrow(\sqrt{2}-1) x+y-1-\frac{1}{\sqrt{2}}+\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)(1-\sqrt{2})=0$