Question

If $P,Q\text{and}R$are the interior angles of trinagles $PQR$,then show that $\tan \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)=\cot \left(\frac{\mathrm{P}}{2}\right)$.

Answer 1

Using the interior angle sum property of a triangle and the trigonometric identity:

$\angle P+\angle Q+\angle R=180°...\left(1\right)$

Dividing (1) by 2,

$\frac{\angle P+\angle Q+\angle R}{2}=90°$

$\Rightarrow \frac{\angle Q+\angle R}{2}=90°-\frac{\angle P}{2}...\left(2\right)$

Now,

$\tan \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)=\cot \left(\frac{\mathrm{P}}{2}\right)$

$LHS=\tan \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)$

$$\begin{gathered} =\tan \left(90-\frac{P}{2}\right)\left[\text{Using equation}\left(2\right)\right] \\ =cot\left(\frac{P}{2}\right)=RHS\left[\because \tan (90°-A)=cotA\right] \end{gathered}$$

Since,

$LHS=RHS$

Hence, the given expression is proved.