<s>[INST] Find tan(C/2) in The triangle ABC . $ABC$ is a triangle.
$\tan\frac{A}{2}  = 0.5$
$\tan\frac{B}{2}  =\frac{1}{3}$.
Find $\tan\frac{C}{2}$.
I tried to find it :
when $A+B+C = 180^{\circ}$ 
So $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$.
How can I complete on these information?
 [/INST] ∴ $\dfrac C2 = 90^0 –(\dfrac A2 + \dfrac B2)$
Then, $\tan (\dfrac {C}{2}) = \cot (\dfrac A2 + \dfrac B2) = \dfrac {1}{\tan (\dfrac A2 + \dfrac B2)} = \dfrac {1 -\tan (\dfrac A2) \tan (\dfrac B2)}{\tan (\dfrac A2) + \tan (\dfrac B2)} = ...$
Added:- It is the application of the compound angle formula $\tan (x + y) = \dfrac {\tan x + \tan y}{1 - \tan x \times \tan y}$.
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