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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If θ+φ=α and θ=k φ (where K>1) , then the value of (θ-φ) is

If θ+ϕ=α\theta+\phi=\alpha and tan⁡θ=ktan⁡ϕ\tan \theta=\mathrm{k} \tan \phi (where K>1)\mathrm{K}>1), then the value of sin⁡(θ−ϕ)\sin (\theta-\phi) is

Options

  1. Aktan⁡ϕ\mathrm{k} \tan \phi
  2. Bsin⁡α\sin \alpha
  3. C\left(\frac{\mathrm{k}-1}{\mathrm{k}+1}\right) \sin \alpha
  4. Dkcos⁡ϕ\mathrm{k} \cos \phi

Correct answer

C. \left(\frac{\mathrm{k}-1}{\mathrm{k}+1}\right) \sin \alpha

Step-by-step solution

We have tan⁡θ=ktan⁡ϕ\tan \theta=\mathrm{k} \tan \phi and θ+ϕ=α\theta+\phi=\alpha \therefore tan⁡θtan⁡ϕ=k1\frac{\tan \theta}{\tan \phi}=\frac{k}{1}By Componendo Dividendo, we gettan⁡θ+tan⁡ϕtan⁡θ−tan⁡ϕ=k+1k−1∴sin⁡θcos⁡θ+sin⁡ϕcos⁡ϕsin⁡θcos⁡θ−sin⁡ϕcos⁡ϕ=k+1k−1∴sin⁡cos⁡ϕ+cos⁡θsin⁡ϕsin⁡θcos⁡ϕ−cos⁡θsin⁡ϕ=k+1k−1∴sin⁡(θ+ϕ)sin⁡(θ−ϕ)=k+1k−1⇒sin⁡αsin⁡(θ−ϕ)=k+1k−1∴sin⁡(θ−ϕ)=k−1k+1(sin⁡α)\begin{aligned}& \frac{\tan \theta+\tan \phi}{\tan \theta-\tan \phi}=\frac{\mathrm{k}+1}{\mathrm{k}-1}\\ & \therefore \frac{\frac{\sin \theta}{\cos \theta}+\frac{\sin \phi}{\cos \phi}}{\frac{\sin \theta}{\cos \theta}-\frac{\sin \phi}{\cos \phi}}=\frac{k+1}{k-1}\\ & \therefore \frac{\sin \cos \phi+\cos \theta \sin \phi}{\sin \theta \cos \phi-\cos \theta \sin \phi}=\frac{k+1}{k-1}\\ & \therefore \frac{\sin (\theta+\phi)}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \Rightarrow \frac{\sin \alpha}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1}\\ & \therefore \sin (\theta-\phi)=\frac{\mathrm{k}-1}{\mathrm{k}+1}(\sin \alpha)\\ & \end{aligned}
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