If cos⁡θ1=2cos⁡θ2,then tan⁡θ1−θ22tan⁡θ1+θ22

If $\cos θ_{1} = 2 \cos θ_{2},$ then $\tan \frac{θ_{1} - θ_{2}}{2} \tan \frac{θ_{1} + θ_{2}}{2}$

A
$\frac{1}{3}$
B
$- \frac{1}{3}$
C
1
D
-1

Solution:

$\tan (\frac{θ_{1} - θ_{2}}{2}) \tan (\frac{θ_{1} + θ_{2}}{2})$

$= \frac{2 \sin (\frac{θ_{1} + θ_{2}}{2}) \sin (\frac{θ_{1} - θ_{2}}{2})}{2 \cos (\frac{θ_{1} + θ_{2}}{2}) \cos (\frac{θ_{1} - θ_{2}}{2})}$

$= \frac{\cos θ_{2} - \cos θ_{1}}{\cos θ_{1} + \cos θ_{2}} = \frac{- 1}{3}$

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