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.