#### Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 1 Maths Textbook Solution.

ANSWER: Equation of tangent, $2x+2y-\pi-4=0$

Equation  of normal ,$2x-2y = \pi$

HINTS:

Differentiate the given equation   with respect to  and to get the slope of the tangent.

GIVEN:

$x=\theta+\sin \theta, y=1+\cos \theta \text { at } \theta=\frac{\pi}{2}$

SOLUTION:

Given as $x=\theta+\sin \theta, y=1+\cos \theta \text { at } \theta=\frac{\pi}{2}$

On differentiating ,

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

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

$m(\text { tangent }) \text { at } \theta=\left(\frac{\pi}{2}\right)-1$

The normal is perpendicular to tangent, therefore , $m_{1} m_{2}=-1$

$m(\text { normal }) \text { at } \theta-\left(\frac{\pi}{2}\right)-1$

The equation of tangent is given by,

\begin{aligned} &y-y_{1}=m(\text { tangent })\left[x-x_{1}\right] \\ &\Rightarrow y-1=-1\left[x-\left(\frac{\pi}{2}\right)-1\right] \end{aligned}

\begin{aligned} &\Rightarrow 2(y-1)+2x -\pi -2=0 \\\Rightarrow &2x+2y-\pi-4 \end{aligned}

The equation of Normal  is given by  ,

\begin{aligned} &y-y_{1}=m(\text { normal })\left[x-x_{1}\right] \\ &\Rightarrow y-1=1\left[x-\left(\frac{\pi}{2}\right)-1\right] \end{aligned}

$2x-2y=\pi$