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Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 1 Maths Textbook Solution.


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ANSWER: Equation of tangent, 2x+2y-\pi-4=0

                 Equation  of normal ,2x-2y = \pi


 Differentiate the given equation   with respect to  and to get the slope of the tangent.


x=\theta+\sin \theta, y=1+\cos \theta \text { at } \theta=\frac{\pi}{2}


Given as x=\theta+\sin \theta, y=1+\cos \theta \text { at } \theta=\frac{\pi}{2}

On differentiating ,

\frac{d x}{d \theta}=1+\cos \theta, \frac{d y}{d \theta}=-\sin \theta

 \therefore \frac{d y}{d x}=-\frac{\sin \theta}{1+\cos \theta}

m(\text { tangent }) \text { at } \theta=\left(\frac{\pi}{2}\right)-1

The normal is perpendicular to tangent, therefore , m_{1} m_{2}=-1

m(\text { normal }) \text { at } \theta-\left(\frac{\pi}{2}\right)-1

The equation of tangent is given by,

\begin{aligned} &y-y_{1}=m(\text { tangent })\left[x-x_{1}\right] \\ &\Rightarrow y-1=-1\left[x-\left(\frac{\pi}{2}\right)-1\right] \end{aligned}

\begin{aligned} &\Rightarrow 2(y-1)+2x -\pi -2=0 \\\Rightarrow &2x+2y-\pi-4 \end{aligned}

The equation of Normal  is given by  ,

\begin{aligned} &y-y_{1}=m(\text { normal })\left[x-x_{1}\right] \\ &\Rightarrow y-1=1\left[x-\left(\frac{\pi}{2}\right)-1\right] \end{aligned}



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