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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If x=a \ θ+ ((θ)/(2))\ and y=a θ , then (d y)/(d x) is equal to

If x=ax=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\} and y=asin⁡θy=a \sin \theta, then dydx\frac{d y}{d x} is equal to

Options

  1. Acot⁡θ\cot \theta
  2. Btan⁡θ\tan \theta
  3. Csin⁡θ\sin \theta
  4. Dcos⁡θ\cos \theta

Correct answer

B. tan⁡θ\tan \theta

Step-by-step solution

Given that, x=a\left(\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right) \text { and }y=asin⁡y=a \sin\thetaOndifferentiatingw.r.t.On differentiating w.r.t.\thetarespectively,wegetrespectively, we getdxdθ=a(−sin⁡θ+1tan⁡(θ2)⋅sec⁡2θ2⋅12)=a(−sin⁡θ+1sin⁡θ)=acos⁡2θsin⁡θ\begin{aligned}\frac{d x}{d \theta}& =a \left(-\sin \theta+\frac{1}{\tan \left(\frac{\theta}{2}\right)} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}\right)\\ & =a\left(-\sin \theta+\frac{1}{\sin \theta}\right)=\frac{a \cos ^2 \theta}{\sin \theta} \end{aligned}andand\frac{d y}{d \theta}=a \cos θ\theta∴dydx=dy/dθdx/dθ=acos⁡θacos⁡2θ/sin⁡θ=tan⁡θ\begin{aligned} \therefore \quad \frac{d y}{d x}& =\frac{d y / d \theta}{d x / d \theta}=\frac{a \cos \theta}{a \cos ^2 \theta / \sin \theta}\\ & =\tan \theta \end{aligned}$
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