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  • Explain solution for RD Sharma Class 12 Chapter 15 Tangent and Normals Exercise Very short Answers Question 17 for maths textbook solution

Explain solution for RD Sharma Class 12 Chapter 15 Tangent and Normals Exercise Very short Answers Question 17 for maths textbook solution.

Answers (1)

Hint :

\begin{aligned} &\text { slope of normal }=\frac{-1}{\text { slope of } \text { tangent }} \\ &\text { i.e., } \frac{-1}{\frac{d y}{d x}} \end{aligned}

Given :

Given curve,

y=cosx \ \ \ \text{ at } (0,1)

We have to write the slope of normal to the given curve at point (0, 1)

Solution :

Here,

            y=cosx \ \ \ \text{ at } (0,1)

On differentiating with respect to x, we get

   \frac{d y}{d x}=-sinx

Now,

\begin{aligned} &\text { slope of } \text { tangent }=\left(\frac{d y}{d x}\right)_{(0,1)}=0 \\ &\text { slope of normal }=\frac{-1}{\text { slope of tangent }} \end{aligned}

                                     =\frac{-1}{0}=\infty

Hence, normal line is x=0.

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