Question

In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]

Answer 1

 In ∆ ABC: 
\[A + B + C = \pi\]
\[ \Rightarrow A + B = \pi - C\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi}{2} - \frac{C}{2}\]
\[\text{ Now, LHS }= \tan\left( \frac{A + B}{2} \right) \]
\[ = \tan\left( \frac{\pi}{2} - \frac{C}{2} \right) \]
\[ = \cot\left( \frac{C}{2} \right) \left[ \because \tan\left( \frac{\pi}{2} - \theta \right) = \cot \theta \right] \]
 = RHS
Hence proved.