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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II1 Mark
MCQ
If $\sin \theta=n \sin (\theta+2 \alpha)$, then $\tan (\theta+\alpha)$ is equal to
  • A
    $\frac{1+n}{2-n} \tan \alpha$
  • B
    $\frac{1-n}{1+n} \tan \alpha$
  • C
    $\tan \alpha$
  • ✓
    $\frac{1+n}{1-n} \tan \alpha$
✓

Answer

Correct option: D.
$\frac{1+n}{1-n} \tan \alpha$
(d) $\frac{1+n}{1-n} \tan \alpha$
Hint:
$\begin{aligned}
& \sin \theta=\mathrm{n} \sin (\theta+2 \alpha) \\
& \frac{\mathrm{n}}{1}=\frac{\sin \theta}{\sin (\theta+2 \alpha)}
\end{aligned}$
By componendo-dividendo, we get
$\begin{aligned}
& \frac{\mathrm{n}+1}{\mathrm{n}-1}=\frac{\sin \theta+\sin (\theta+2 \alpha)}{\sin \theta-\sin (\theta+2 \alpha)} \\
& =\frac{2 \sin (\theta+\alpha) \cos \alpha}{-2 \cos (\theta+\alpha) \sin \alpha} \\
& \tan (\theta+\alpha)=-\left(\frac{\mathrm{n}+1}{\mathrm{n}-1}\right) \tan \alpha=\frac{1+\mathrm{n}}{1-\mathrm{n}} \tan \alpha \\
\end{aligned}$

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