In fig, find tan P – cot R.

Solution:

In figure, by the Pythagoras Theorem

$\mathrm{QR}^{2}=\mathrm{PR}^{2}-\mathrm{PQ}^{2}=(13)^{2}-(12)^{2}=25$

$\Rightarrow \mathrm{QR}=\sqrt{25}=5 \mathrm{~cm}$

In $\triangle \mathrm{PQR}$ right angled at $\mathrm{Q}, \mathrm{QR}=5 \mathrm{~cm}$ is side opposite to the angle $\mathrm{P}$ and $\mathrm{PQ}=12 \mathrm{~cm}$ is side adjacent to the angle P.

Therefore, $\tan \mathrm{P}=\frac{\mathbf{Q R}}{\mathbf{P Q}}=\frac{\mathbf{5}}{\mathbf{1 2}}$.

Now, $Q R=5 \mathrm{~cm}$ is side adjacent to the angle $R$ and $P Q=12 \mathrm{~cm}$ is side opposite to the angle $\mathrm{R}$.

Therefore, $\cot \mathrm{R}=\frac{\mathbf{Q R}}{\mathbf{P Q}}=\frac{\mathbf{5}}{\mathbf{1 2}}$

Hence, $\tan P-\cot R=\frac{\mathbf{5}}{\mathbf{1 2}}-\frac{\mathbf{5}}{\mathbf{1 2}}=\mathbf{0}$