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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.

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

Answers (1)

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).

Solution:

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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