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

Given:

\( \angle A \) and \( \angle P \) are acute angles such that \( \tan A=\tan P \).

To do:

We have to show that \( \angle A=\angle P \).

Solution:  

Let, in a triangle $APC$ right angled at $C$, $tan\ A = tan\ P$.

We know that,

In a right-angled triangle $APC$ with right angle at $C$,

By trigonometric ratios definitions,

$tan\ A=\frac{Opposite}{Adjacent}=\frac{PC}{AC}$

$tan\ P=\frac{Opposite}{Adjacent}=\frac{AC}{PC}$

This implies,

$\tan A=\tan P$

$\Rightarrow \frac{PC}{AC}=\frac{AC}{PC}$

 $\Rightarrow PC \times PC=AC \times AC$

 $\Rightarrow PC^2 =AC^2 $

$\Rightarrow PC = AC $

We know that,

Angles opposite to equal sides are equal in a triangle.

Therefore,

$\angle A=\angle P$

Hence proved.