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

# If $\mathrm{cos}{\mathrm{\theta }}_{1}=2\mathrm{cos}{\mathrm{\theta }}_{2},$then $\mathrm{tan}\frac{{\mathrm{\theta }}_{1}-{\mathrm{\theta }}_{2}}{2}\mathrm{tan}\frac{{\mathrm{\theta }}_{1}+{\mathrm{\theta }}_{2}}{2}$

1. A

$\frac{1}{3}$

2. B

$-\frac{1}{3}$

3. C

1

4. D

-1

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### Solution:

$\mathrm{tan}\left(\frac{{\mathrm{\theta }}_{1}-{\mathrm{\theta }}_{2}}{2}\right)\mathrm{tan}\left(\frac{{\mathrm{\theta }}_{1}+{\mathrm{\theta }}_{2}}{2}\right)$

$=\frac{2\mathrm{sin}\left(\frac{{\theta }_{1}+{\theta }_{2}}{2}\right)\mathrm{sin}\left(\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)}{2\mathrm{cos}\left(\frac{{\theta }_{1}+{\theta }_{2}}{2}\right)\mathrm{cos}\left(\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)}$

$=\frac{\mathrm{cos}{\mathrm{\theta }}_{2}-\mathrm{cos}{\mathrm{\theta }}_{1}}{\mathrm{cos}{\mathrm{\theta }}_{1}+\mathrm{cos}{\mathrm{\theta }}_{2}}=\frac{-1}{3}$