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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2017 · Mathematics · Trigonometric Ratios & Identities

Angle α is divided into two parts A and B such that A-B=x and A:...

Angle α\alpha is divided into two parts A and B such that A−B=xA-B=x and tan⁡A:tan⁡B=p:q\tan A: \tan B=p: q. The value of sin⁡x\sin x is equal to

Options

  1. A(p+q)sin⁡αp−q\frac{(\mathrm{p}+\mathrm{q}) \sin \alpha}{\mathrm{p}-\mathrm{q}}
  2. Bpsin⁡αp+q\frac{\mathrm{p} \sin \alpha}{\mathrm{p}+\mathrm{q}}
  3. Cpsin⁡αp−q\frac{\mathrm{p} \sin \alpha}{\mathrm{p}-\mathrm{q}}
  4. D(p−q)sin⁡αp+q\frac{(\mathrm{p}-\mathrm{q}) \sin \alpha}{\mathrm{p}+\mathrm{q}}

Correct answer

D. (p−q)sin⁡αp+q\frac{(\mathrm{p}-\mathrm{q}) \sin \alpha}{\mathrm{p}+\mathrm{q}}

Step-by-step solution

A−B=x\mathrm{A}-\mathrm{B}=\mathrm{x} and tan⁡A:tan⁡\tan \mathrm{A}: \tan\mathrm{B}=\mathrm{p}:q: \mathrm{q} Also, given α=A+BA−B=x\alpha=\mathrm{A}+\mathrm{B}\mathrm{A}-\mathrm{B}=\mathrm{x} and α=A+B⇒A=x+α2andB=α−x2\alpha=\mathrm{A}+\mathrm{B}\Rightarrow \mathrm{A}=\frac{\mathrm{x}+\alpha}{2} and \mathrm{B}=\frac{\alpha-\mathrm{x}}{2} Now, tan⁡Atan⁡B=tan⁡(x+α2)tan⁡(α−x2)=pq\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=\frac{\tan \left(\frac{\mathrm{x}+\alpha}{2}\right)}{\tan \left(\frac{\alpha-\mathrm{x}}{2}\right)}=\frac{\mathrm{p}}{\mathrm{q}} (Given) ⇒2sin⁡(α+x2)cos⁡(α−x2)2cos⁡(α+x2)sin⁡(α−x2)=pq⇒sin⁡α+sin⁡xsin⁡α−sin⁡x=pq\begin{aligned} & \Rightarrow \frac{2 \sin \left(\frac{\alpha+x}{2}\right) \cos \left(\frac{\alpha-x}{2}\right)}{2 \cos \left(\frac{\alpha+x}{2}\right) \sin \left(\frac{\alpha-x}{2}\right)}=\frac{p}{q} \\ & \Rightarrow \frac{\sin \alpha+\sin x}{\sin \alpha-\sin x}=\frac{p}{q}\end{aligned}\Rightarrow \frac{\sin \alpha}{\sin x}=\frac{p+q}{p-q} \Rightarrow \sin x=\frac{(\sin \alpha)(p-q)}{p+q}$
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