#### Explain solution RD Sharma class 12 Chapter 15 Tangents and Normals exercise Fill in the blanks question 1

Equation of normal, $y=-x$

Hint:

Use equation of normal formula,

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

Given:

Here given the curve , $y=\tan x$

To find:

We have to find the equation of normal to the curve at $\left ( 0,0 \right )$

Solution:

Here given,

$y=\tan x$

Differentiating both side with respect to $x$ , we get

\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=\sec ^{2} x \\\\ &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(0,0)}=\sec ^{2}(0)=1 \\\\ &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(0.0)}=1 \end{aligned}

Slope of normal at  $\left ( 0,0 \right )=\frac{-1}{\frac{dy}{dx}}$

$=-1$

Now, equation of normal to the curve  $y=\tan x$  at  $\left ( 0,0 \right )$  is

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

$\Rightarrow$                $(y-0)=(-1) x$

$\Rightarrow$                $y=-x$

Hence, this is the required equation of normal.