CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Differentiating it with respect to $\theta$

$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta).....(\text{i}) $

And, $\text{y}=\text{a}(1-\cos\theta)$

Differentiating it with respect to $\theta$, 

$ \frac{\text{dy}}{\text{d}\theta}=\text{a}(\theta+\sin\theta)$

and

$\frac{\text{dy}}{\text{d}\theta}=\text{a}\sin\theta...(\text{ii}) $

Using equation (i) and (ii),

$=\frac{\text{a}\sin\theta}{\text{a}(1-\cos\theta)} $

$=\frac{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\sin^{2}\theta}{2}}, \begin{Bmatrix} \text{Since, }1-\cos\theta=\frac{2\sin^{2\theta}}{2}\\\frac{2\sin\theta}{2}\frac{\cos\theta}{2}=\sin\theta \end{Bmatrix}$

$=\frac{\text{dy}}{\text{dx}}=\frac{\tan}{2}$