Question.
Given $\sec \theta=\frac{\mathbf{1 3}}{\mathbf{1 2}}$, calculate all other trigonometric ratios.
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,
$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$
$(13 k)^{2}=A B^{2}+(12 k)^{2}$
$\mathrm{AB}^{2}=169 \mathrm{k}^{2}-144 \mathrm{k}^{2}$
$\mathrm{AB}=\sqrt{25 \mathrm{k}^{2}}=5 \mathrm{k}$
$\sin \theta=\frac{\mathbf{A B}}{\mathbf{A C}}=\frac{5 \mathbf{k}}{\mathbf{1 3 k}}=\frac{\mathbf{5}}{\mathbf{1 3}}$
$\cos \theta=\frac{\mathbf{B C}}{\mathbf{A C}}=\frac{12 \mathbf{k}}{\mathbf{1 3 k}}=\frac{12}{13}$
$\tan \theta=\frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{5 k}}{\mathbf{1 2 k}}=\frac{\mathbf{5}}{\mathbf{1 2}}$
$\cot \theta=\frac{\mathbf{B C}}{\mathbf{A B}}=\frac{12 \mathbf{k}}{5 \mathbf{k}}=\frac{12}{5}$
$\operatorname{cosec} \theta=\frac{\mathbf{A C}}{\mathbf{A B}}=\frac{\mathbf{1 3 k}}{\mathbf{5 k}}=\frac{\mathbf{1 3}}{\mathbf{5}}$
$\sec \theta=\frac{13}{12}$
$\Rightarrow \frac{A C}{B C}=\frac{13}{12}$
By Pythagoras Theorem,
$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$
$(13 k)^{2}=A B^{2}+(12 k)^{2}$
$\mathrm{AB}^{2}=169 \mathrm{k}^{2}-144 \mathrm{k}^{2}$
$\mathrm{AB}=\sqrt{25 \mathrm{k}^{2}}=5 \mathrm{k}$
$\sin \theta=\frac{\mathbf{A B}}{\mathbf{A C}}=\frac{5 \mathbf{k}}{\mathbf{1 3 k}}=\frac{\mathbf{5}}{\mathbf{1 3}}$
$\cos \theta=\frac{\mathbf{B C}}{\mathbf{A C}}=\frac{12 \mathbf{k}}{\mathbf{1 3 k}}=\frac{12}{13}$
$\tan \theta=\frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{5 k}}{\mathbf{1 2 k}}=\frac{\mathbf{5}}{\mathbf{1 2}}$
$\cot \theta=\frac{\mathbf{B C}}{\mathbf{A B}}=\frac{12 \mathbf{k}}{5 \mathbf{k}}=\frac{12}{5}$
$\operatorname{cosec} \theta=\frac{\mathbf{A C}}{\mathbf{A B}}=\frac{\mathbf{1 3 k}}{\mathbf{5 k}}=\frac{\mathbf{1 3}}{\mathbf{5}}$
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