Let x = \theta + \cos\theta, \; y = \theta - \sin\theta represents parametric form of a curve, then slope of tangent at a general point is ?

  • Option 1)

    \frac{1 - \cos\theta}{1 - \sin\theta}

  • Option 2)

    \frac{1 + \cos\theta}{1 - \sin\theta}

  • Option 3)

    \frac{1 - \cos\theta}{1 + \sin\theta}

  • Option 4)

    \frac{1 + \cos\theta}{1 + \sin\theta}

 

Answers (1)

As we have learned

Slope of tangent for parametric form -

M_{T}=\frac{dy}{dt}.\frac{dt}{dx}

      
        =\frac{f'_{y}}{f'_{x}}

and\:find \:M_{T}\:at\:(t)

- wherein

Where\:\:y=f(t)\:\:\:and\:\:\:x=f(t)

 

 Slope of tangent = \frac{dy}{dx} =\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}= \frac{1- \cos \theta }{1- \sin \theta }

 

 

 

 

 

 


Option 1)

\frac{1 - \cos\theta}{1 - \sin\theta}

Option 2)

\frac{1 + \cos\theta}{1 - \sin\theta}

Option 3)

\frac{1 - \cos\theta}{1 + \sin\theta}

Option 4)

\frac{1 + \cos\theta}{1 + \sin\theta}

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