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Solution for Find the Equation of the Tangent to the Curve X = θ + Sin θ, Y = 1 + Cos θ at θ = π/4 ? - CBSE (Science) Class 12 - Mathematics

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Question

Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4 ?

Solution

\[x = \theta + \sin \theta \text { and } y = 1 + \cos \theta\]

\[\frac{dx}{d\theta} = 1 + \cos \theta \text { and } \frac{dy}{d\theta} = - \sin \theta\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- \sin \theta}{1 + \cos \theta}\]

\[\text { Slope of tangent }= \left( \frac{dy}{dx} \right)_\theta = \frac{\pi}{4} =\frac{- \sin \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}}=\frac{\frac{- 1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}=\frac{-1}{\sqrt{2} + 1}=\frac{-1}{\sqrt{2} + 1}\times\frac{\sqrt{2} - 1}{\sqrt{2} - 1}=1 - \sqrt{2}\]

\[\left( x_1 , y_1 \right) = \left( \frac{\pi}{4} + \sin\frac{\pi}{4}, 1 + \cos \frac{\pi}{4} \right) = \left( \frac{\pi}{4} + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}} \right)\]

\[\text { Equation of tangent is },\]

\[y - y_1 = m\left( x - x_1 \right)\]

\[ \Rightarrow y - \left( 1 + \frac{1}{\sqrt{2}} \right) = \left( 1 - \sqrt{2} \right)\left[ x - \left( \frac{\pi}{4} + \frac{1}{\sqrt{2}} \right) \right]\]

\[ \Rightarrow y - 1 - \frac{1}{\sqrt{2}} = \left( 1 - \sqrt{2} \right)\left[ x - \frac{\pi}{4} - \frac{1}{\sqrt{2}} \right]\]

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Solution Find the Equation of the Tangent to the Curve X = θ + Sin θ, Y = 1 + Cos θ at θ = π/4 ? Concept: Tangents and Normals.
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