If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.

Solution:

Given: $\tan A=\tan P$

To show: $\angle A=\angle P$

Consider two right angled triangles $\mathrm{ABC}$ and $\mathrm{PQR}$ such that $\tan A=\tan P$

Therefore we have,

$\tan A=\frac{B C}{A B}$ and $\tan P=\frac{Q R}{P Q}$

Since it is given that $\tan A=\tan P$

Therefore,

$\frac{B C}{A B}=\frac{Q R}{P Q}$

Now by interchanging position of AB and QR by cross multiplication

We get,

$\frac{B C}{Q R}=\frac{A B}{P Q}$

Let $\frac{B C}{Q R}=\frac{A B}{P Q}=k$ (say)....(1)

Now by cross multiplication

$B C=k Q R$ and $A B=k P Q \ldots \ldots(2)$

Now by using Pythagoras theorem in triangles ABC and PQR

We have,

$A C^{2}=A B^{2}+B C^{2}$ and $P R^{2}=P Q^{2}+Q R^{2}$

Therefore

$A C=\sqrt{A B^{2}+B C^{2}}$ and $P R=\sqrt{P Q^{2}+Q R^{2}}$

Now $\frac{A C}{P R}=\frac{\sqrt{A B^{2}+B C^{2}}}{\sqrt{P Q^{2}+Q R^{2}}}$

Now using equation (2)

We get,

$\frac{A C}{P R}=\frac{\sqrt{(k P Q)^{2}+(k Q R)^{2}}}{\sqrt{P Q^{2}+Q R^{2}}}$

$\frac{A C}{P R}=\frac{\sqrt{k^{2} P Q^{2}+k^{2} Q R^{2}}}{\sqrt{P Q^{2}+Q R^{2}}}$

Now by taking $k^{2}$ common

We get,

$\frac{A C}{P R}=\frac{\sqrt{k^{2}\left(P Q^{2}+Q R^{2}\right)}}{\sqrt{P Q^{2}+Q R^{2}}}$

Therefore,

$\frac{A C}{P R}=\frac{k \sqrt{\left(P Q^{2}+Q R^{2}\right)}}{\sqrt{P Q^{2}+Q R^{2}}}$

Now $\sqrt{P Q^{2}+Q R^{2}}$ gets cancelled

Therefore,

$\frac{A C}{P R}=k$....(3)

From (1) and (3)

$\frac{B C}{Q R}=\frac{A B}{P Q}=\frac{A C}{P R}=k$

Therefore, $\triangle A B C \sim \Delta P Q R$

Hence, $\angle A=\angle P$