Question

If A, B and C are the angles of a $∆$ABC, prove that $\tan \left(\frac{\mathrm{C}+\mathrm{A}}{2}\right)=\cot \frac{\mathrm{B}}{2}$.

Answer 1

$$\begin{gathered} \mathrm{In}∆\mathrm{ABC}, \\ \mathrm{A}+\mathrm{B}+\mathrm{C}=180° \\ \Rightarrow \mathrm{A}+\mathrm{C}=180°-\mathrm{B}.....\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ \mathrm{LHS}=\tan \left(\frac{\mathrm{C}+\mathrm{A}}{2}\right) \\ =\tan \left(\frac{180°-\mathrm{B}}{2}\right)\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ =\tan \left(90°-\frac{\mathrm{B}}{2}\right) \\ =\cot \frac{\mathrm{B}}{2} \\ =\mathrm{RHS} \end{gathered}$$