Q62 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$, then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$

  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{4}$
  • Correct answer
    C
    $\frac{1}{3}$
  • D
    $\frac{1}{8}$

Solution

Given, $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$ $\therefore \quad \frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2}{1}$ Now, using componendo and dividendo, $\Rightarrow \frac{\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha-\beta}{2}\right)-\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2+1}{2-1}$ $\Rightarrow \quad \frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=3 \Rightarrow \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1}{3}$

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

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