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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}=2\cos\theta-\cot2\theta$ and $\text{y}=2\sin\theta-\sin2\theta,$ prove that $\frac{\text{dy}}{\text{dx}}=\tan\big(\frac{3\theta}{2}\big)$
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Answer

Here, $\text{x}=2\cos\theta-\cos2\theta$
Diffierentiating it with respect to $\theta$ using chain rule, 
$\frac{\text{dx}}{\text{d}\theta}=2(-\sin\theta)-(-\sin2\theta)\frac{\text{d}}{\text{d}\theta}(2\theta)$
$=-2\sin\theta+2\sin2\theta$
$\frac{\text{dx}}{\text{d}\theta}=2(\sin2\theta-\sin\theta)...(\text{i})$
And, $\text{y}=2\sin\theta-\sin\theta$
Differentiating it with respect to $\theta$ using chain rule, 
$\frac{\text{dt}}{\text{d}\theta}=2\cos\theta-\cos2\theta\frac{\text{d}}{\text{d}\theta}(2\theta)$
$=2\cos\theta-\cos2\theta(2)$
$=2\cos\theta-\cos2\theta(2)$
$\frac{\text{dy}}{\text{d}\theta}=2(\cos\theta-\cos2\theta)...(\text{ii})$
Dividing equation (ii) by equation (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{2(\cos\theta-\cos2\theta)}{2(\sin2\theta-\sin\theta)}$
$=\frac{(\cos\theta-\cos2\theta)}{(\sin2\theta-\sin\theta)}$
$\frac{\text{dy}}{\text{dx}}=\frac{-2\sin\big(\frac{\theta+2\theta}{2}\big)\sin\big(\frac{\theta-2\theta}{2}\big)}{2\cos\big(\frac{2\theta+\theta}{2}\big)\sin\big(2\theta-\theta\big)}$
$\Big[\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A+B}}{2}\Big)\Big]$ 
$\Rightarrow-\frac{\sin\big(\frac{3\theta}{2}\big)\big(\sin\big(\frac{-\theta}{2}\big)\big)}{\cos\big(\frac{3\theta}{2}\big)\sin\big(\frac{\theta}{2}\big)}$
$\Rightarrow-\frac{\sin\big(\frac{3\theta}{2}\big)\big(-\sin\frac{-\theta}{2}\big)}{\cos\big(\frac{3\theta}{2}\big)\sin\big(\frac{\theta}{2}\big)}$
$\Rightarrow\frac{\sin\big(\frac{3\theta}{2}\big)}{\cos\big(\frac{3\theta}{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\tan\big(\frac{3\theta}{2}\big)$

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