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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
Given that $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$, then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$ is equal to
  • A
    $\frac{1}{2}$
  • ✓
    $\frac{1}{3}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{8}$
✓

Answer

Correct option: B.
$\frac{1}{3}$
(B)
$\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$
$\Rightarrow \cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}$
$=2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}-2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}$
$\Rightarrow 3 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}=\cos \frac{\alpha}{2} \cos \frac{\beta}{2}$
$\Rightarrow \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1}{3}$

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