First slide

Multiple and sub- multiple Angles

Question

If $\cos θ = \cos α c o s β,$ then $\tan \frac{θ + α}{2} \tan \frac{θ - α}{2}$ is equal to

Moderate

A

$\tan^{2} (α / 2)$
B

$\tan^{2} (β / 2)$
C

$\tan^{2} (θ / 2)$
D

$\cot^{2} (β / 2)$

Solution

$\tan \frac{θ + α}{2} \tan \frac{θ - α}{2}$
$= \frac{\tan^{2} (θ / 2) - \tan^{2} (α / 2)}{1 - \tan^{2} (θ / 2) \tan^{2} (α / 2)}$
$= \frac{\frac{1 - \cos θ}{1 + \cos θ} - \frac{1 - \cos α}{1 + \cos α}}{1 - \frac{1 - \cos θ}{1 + \cos θ} \cdot \frac{1 - \cos α}{1 + \cos α}}$
$= \frac{2 (\cos α - c o s θ)}{2 (\cos α + \cos θ)} = \frac{\cos α (1 - \cos β)}{\cos α (1 + \cos β)} = \tan^{2} \frac{β}{2}$

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