# If y = 1 - cos θ, x = 1 - sinθ then $\frac{{dy}}{{dx}}$at θ = $\frac{\pi}{4}$is

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If y = 1 - cos θ, x = 1 - sinθ then $\frac{{dy}}{{dx}}$at θ = $\frac{\pi}{4}$is
1. - 1
2. 1
3. 1/2
4. $\frac{1}{\sqrt2}$
5. None of these

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Correct Answer - Option 1 : - 1

CONCEPT:

• $\frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x$
• $\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$

Differentiation of parametric functions:

If x = f(t), y = g(t), where t is a parameter, then $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{g'\left( t \right)}}{{f'\left( t \right)}}$

CALCULATION:

Given: y = 1 - cosθ, x = 1 - sin θ

Here, we have to find $\frac{{dy}}{{dx}}$ at θ = π/4

As we can see that, x and y are functions of θ

We also know that, if x = f(t), y = g(t), where t is a parameter, then $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{g'\left( t \right)}}{{f'\left( t \right)}}$

Let's differentiate x and y w.r.t θ

⇒ $\frac{{dy}}{{d\theta }} = \frac{{d\left( {1 - \cos \theta } \right)}}{{d\theta }} = \sin \theta$       ---(1)

⇒ $\frac{{dx}}{{d\theta }} = \frac{{d\left( {1 - \sin \theta } \right)}}{{d\theta }} = \; - \cos \theta$       ---(2)

Now from (1) and (2) we can say that, $\frac{{dy}}{{dx}} = \; - \tan \theta$

⇒ ${\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {\theta = \frac{\pi }{4}} \right)}} = \; - \tan \frac{\pi }{4} = \; - 1$

Hence, option A is the correct answer.