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

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Equation of normal, y=-x


Use equation of normal formula,

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


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 )


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}}


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.


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