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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

If p (θ-30^ ^ )=q (θ+120^ ^ ), then what is ((p+q))/((p-q)) equal...

If ptan⁡(θ−30∘)=qtan⁡(θ+120∘),thenwhatis(p+q)(p−q)equalto?equalto?p \tan \left(\theta-30^{{}^{\circ}}\right)=q \tan \left(\theta+120^{{}^{\circ}}\right), then what is \frac{(p+q)}{(p-q)} equal to? equal to?

Options

  1. Asin⁡2θ\sin 2 \theta
  2. Bcos⁡2θ\cos 2 \theta
  3. C2sin⁡2θ2 \sin 2 \theta
  4. D2cos⁡2θ2 \cos 2 \theta

Correct answer

D. 2cos⁡2θ2 \cos 2 \theta

Step-by-step solution

We have. ptan⁡(θ−30∘)=qtan⁡(θ+120∘)⇒pq=tan⁡(θ+120∘)tan⁡(θ−30∘)=sin⁡(θ+120∘)cos⁡(θ−30∘)cos⁡(θ+120∘)sin⁡(θ−30∘)\begin{aligned} & p \tan \left(\theta-30^{{}^{\circ}}\right)=q \tan \left(\theta+120^{{}^{\circ}}\right) \\ & \Rightarrow \frac{p}{q}=\frac{\tan \left(\theta+120^{{}^{\circ}}\right)}{\tan \left(\theta-30^{{}^{\circ}}\right)} \\ & =\frac{\sin \left(\theta+120^{{}^{\circ}}\right) \cos \left(\theta-30^{{}^{\circ}}\right)}{\cos \left(\theta+120^{{}^{\circ}}\right) \sin \left(\theta-30^{{}^{\circ}}\right)} \end{aligned} Using componendo and dividendo rule, ⇒p+qp−q=sin⁡(θ+120∘)cos⁡(θ−30∘)+cos⁡(θ+120∘)sin⁡(θ−30∘)sin⁡(θ+120∘)cos⁡(θ−30∘)−cos⁡(θ+120∘)sin⁡(θ−30∘)⇒p+qp−q=sin⁡[(θ+120∘)+(θ−30∘)]sin⁡[(θ+120∘)−(θ−30∘)]=sin⁡(90∘+2θ)sin⁡(150∘)⇒p+qp−q=sin⁡(90∘+2θ)(12)=2cos⁡2θ\begin{gathered} \Rightarrow \frac{p+q}{p-q}=\frac{\sin \left(\theta+120^{{}^{\circ}}\right) \cos \left(\theta-30^{{}^{\circ}}\right)+\cos \left(\theta+120^{{}^{\circ}}\right) \sin \left(\theta-30^{{}^{\circ}}\right)}{\sin \left(\theta+120^{{}^{\circ}}\right) \cos \left(\theta-30^{{}^{\circ}}\right)-\cos \left(\theta+120^{{}^{\circ}}\right) \sin \left(\theta-30^{{}^{\circ}}\right)} \\ \Rightarrow \frac{p+q}{p-q}=\frac{\sin \left[\left(\theta+120^{{}^{\circ}}\right)+\left(\theta-30^{{}^{\circ}}\right)\right]}{\sin \left[\left(\theta+120^{{}^{\circ}}\right)-\left(\theta-30^{{}^{\circ}}\right)\right]}=\frac{\sin \left(90^{{}^{\circ}}+2 \theta\right)}{\sin \left(150^{{}^{\circ}}\right)} \\ \Rightarrow \frac{p+q}{p-q}=\frac{\sin \left(90^{{}^{\circ}}+2 \theta\right)}{\left(\frac{1}{2}\right)}=2 \cos 2 \theta \end{gathered}
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