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If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$, then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$
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The correct answer is:
$\frac{1}{3}$
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,
$\Rightarrow \frac{\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha-\beta}{2}\right)-\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2+1}{2-1}$
$\Rightarrow \quad \frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=3 \Rightarrow \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1}{3}$
$\therefore \quad \frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2}{1}$
Now, using componendo and dividendo,
$\Rightarrow \frac{\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha-\beta}{2}\right)-\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2+1}{2-1}$
$\Rightarrow \quad \frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=3 \Rightarrow \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1}{3}$
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