Question
In a $\triangle\text{ABC},$ Prove that:
$\tan\frac{\text{A+B}}{2}=\cot\frac{\text{C}}{2}$

Answer

We have:
$\text{A + B + C} = \pi$ $\big(\because$ sum of 3 angle of a triangle is $\pi=180^\circ\big)$
$\Rightarrow\text{A+B}=\pi-\text{C}$
$\Rightarrow\frac{\text{A+B}}{2}=\frac{\pi-\text{C}}{2}$
$\Rightarrow\frac{\text{A+B}}{2}=\frac{\pi}{2}-\frac{\text{C}}{2}$
$\Rightarrow\tan\Big(\frac{\text{A+B}}{2}\Big)=\tan\Big(\frac{\pi}{2}-\frac{\text{C}}{2}\Big)$
$=\cot\frac{\text{C}}{2}$ $\Big(\because\tan\Big(\frac{\pi}{2}-\theta\Big)=\cot\theta\Big)$
Hence, $\tan\Big(\frac{\text{A+B}}{2}\Big)=\tan\Big(\frac{\pi}{2}-\frac{\text{C}}{2}\Big)$
$\text{Proved}$

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