Q184 / 270 · NDA 2017 (Phase 2) · Mathematics

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

  • A
    $\frac{(\mathrm{p}+\mathrm{q}) \sin \alpha}{\mathrm{p}-\mathrm{q}}$
  • B
    $\frac{\mathrm{p} \sin \alpha}{\mathrm{p}+\mathrm{q}}$
  • C
    $\frac{\mathrm{p} \sin \alpha}{\mathrm{p}-\mathrm{q}}$
  • Correct answer
    D
    $\frac{(\mathrm{p}-\mathrm{q}) \sin \alpha}{\mathrm{p}+\mathrm{q}}$

Solution

$\mathrm{A}-\mathrm{B}=\mathrm{x}$ and $\tan \mathrm{A}: \tan \mathrm{B}=\mathrm{p}: \mathrm{q}$ Also, given $\alpha=\mathrm{A}+\mathrm{B}$ $\mathrm{A}-\mathrm{B}=\mathrm{x}$ and $\alpha=\mathrm{A}+\mathrm{B}$ $\Rightarrow \mathrm{A}=\frac{\mathrm{x}+\alpha}{2}$ and $\mathrm{B}=\frac{\alpha-\mathrm{x}}{2}$ Now, $\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=\frac{\tan \left(\frac{\mathrm{x}+\alpha}{2}\right)}{\tan \left(\frac{\alpha-\mathrm{x}}{2}\right)}=\frac{\mathrm{p}}{\mathrm{q}}$ (Given) $\begin{aligned} & \Rightarrow \frac{2 \sin \left(\frac{\alpha+x}{2}\right) \cos \left(\frac{\alpha-x}{2}\right)}{2 \cos \left(\frac{\alpha+x}{2}\right) \sin \left(\frac{\alpha-x}{2}\right)}=\frac{p}{q} \\ & \Rightarrow \frac{\sin \alpha+\sin x}{\sin \alpha-\sin x}=\frac{p}{q}\end{aligned}$ $\Rightarrow \frac{\sin \alpha}{\sin x}=\frac{p+q}{p-q} \Rightarrow \sin x=\frac{(\sin \alpha)(p-q)}{p+q}$

Hence, the correct answer is option: 'D'.

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