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Explain solution for RD Sharma Class 12 Chapter 15 Tangent and Normals Exercise Very short Answers Question 18 for maths textbook solution.

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Answer : y=x is required equation of tangent

Hint :

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

Given :

Given curve,

y=\sin x

We have to write the equation of tangent drawn to the given curve at point (0,0).


y=\sin x

Differentiating both sides with respect to x, we get

\begin{aligned} &\frac{d y}{d x}=\cos x \\ &\left(\frac{d y}{d x}\right)_{(0,0)}=\cos 0=1 \end{aligned}

Hence the equation of tangent at (0,0),

\begin{aligned} &(y-0)=1(x-0) \\ &y=x \end{aligned}

Hence, y=x is the required equation.

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