#### Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 10 maths text book solution.

Answer : Equation of normal $=2x = \pi$

Hint : Use equation of normal,

$y-y_{1}=\frac{1}{\frac{d y}{d x}}\left(x-x_{1}\right)$

Given :

Here given that the curve

$y=x+\sin x \cos x$

We have to write the equation of normal to the given curve at $x=\frac{\pi }{2}$

Solution:

We have,

$y=x+\sin x \cos x$

On differentiating both sides with respect to  $'x'$, we get

$\frac{d y}{d x}=1+\cos ^{2} x-\sin ^{2} x$

Now we know that,

\begin{aligned} &\text { slope of tangent }=\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}} \\ &=1+\cos ^{2}\left(\frac{\pi}{2}\right)-\sin ^{2}\left(\frac{\pi}{2}\right) \\ &=1+0-1 \\ &=0 \end{aligned}

Now,

When $x=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}+\sin \frac{\pi}{2} \cos \frac{\pi}{2}$

$\Rightarrow y=\frac{\pi }{2}$

$\therefore \quad\left(x_{1}, y_{1}\right)=\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$

So, equation of normal

\begin{aligned} &y-y_{1}=\frac{-1}{\text { slope of tangent }}\left(x-x_{1}\right) \\ &y-\frac{\pi}{2}=\frac{-1}{0}\left(x-\frac{\pi}{2}\right) \end{aligned}

$\begin{array}{ll} \Rightarrow \quad & x=\frac{\pi}{2} \\ \Rightarrow & 2 x=\pi \end{array}$

Hence the equation of the given curve at $x=\frac{\pi}{2} \text { is } 2 x=\pi$.