Given $\sec \theta=\frac{\mathbf{1 3}}{\mathbf{1 2}}$, calculate all other trigonometric ratios.

Solution:

$\sec \theta=\frac{13}{12}$

$\Rightarrow \frac{A C}{B C}=\frac{13}{12}$

By Pythagoras Theorem,



$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$

$(13 k)^{2}=A B^{2}+(12 k)^{2}$

$\mathrm{AB}^{2}=169 \mathrm{k}^{2}-144 \mathrm{k}^{2}$

$\mathrm{AB}=\sqrt{25 \mathrm{k}^{2}}=5 \mathrm{k}$

$\sin \theta=\frac{\mathbf{A B}}{\mathbf{A C}}=\frac{5 \mathbf{k}}{\mathbf{1 3 k}}=\frac{\mathbf{5}}{\mathbf{1 3}}$

$\cos \theta=\frac{\mathbf{B C}}{\mathbf{A C}}=\frac{12 \mathbf{k}}{\mathbf{1 3 k}}=\frac{12}{13}$

$\tan \theta=\frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{5 k}}{\mathbf{1 2 k}}=\frac{\mathbf{5}}{\mathbf{1 2}}$

$\cot \theta=\frac{\mathbf{B C}}{\mathbf{A B}}=\frac{12 \mathbf{k}}{5 \mathbf{k}}=\frac{12}{5}$

$\operatorname{cosec} \theta=\frac{\mathbf{A C}}{\mathbf{A B}}=\frac{\mathbf{1 3 k}}{\mathbf{5 k}}=\frac{\mathbf{1 3}}{\mathbf{5}}$