If $A, B$ and $C$ are interior angles of a triangle $A B C$ , then show that tan $(\frac{A + B}{2}) = cot \frac{C}{2}$

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Solution

$A + B + C = 180^{0} ∴ t a n ((\frac{A + B}{2})) = t a n ((\frac{180^{0} - C}{2})) = t a n (90^{0} - (\frac{C}{2})) = c o t ((\frac{C}{2})) = R H S$
Hence proved!

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