Question

If $\angle A$ and $\angle P$ are acute angles such that tan A = tan P, then show that $\angle A=\angle P$

Answer 1

Consider $\Delta$ ABC is a right angled triangle

According to the question,

tan A = tan P

$\therefore$ tan A = $\frac{BC}{AB}$ and tan P = $\frac{AB}{BC}$

$\Rightarrow$ $\frac{BC}{AB}$ = $\frac{AB}{BC}$

$\Rightarrow$ $B{C}^2$ = $A{B}^2$

$\Rightarrow$ BC = AB

So, tan A = $\frac{BC}{AB}$ = $\frac{AB}{AB}$ = 1

$\Rightarrow$ tan A = 1

$\therefore$ A = ${45}^{\circ }$

Same way, tan P = 1 and P = ${45}^{\circ }$

Hence, $\angle P$ = $\angle A$