Question
In the given figure, $A C=13 \ cm , B C=12 \ cm$ and $\angle B=90^{\circ}$. Without using tables, find the values of: $\frac{\cos A-\sin A}{\cos A+\sin A}$
✓
Answer
$\triangle A B C$ is a right$-$angled triangle.
$ \therefore A C^2=A B^2+B C^2$
$\Rightarrow A B^2$
$=A C^2-B C^2$
$=13^2-12^2$
$=169-144$
$=25$
$\Rightarrow A B=5 \ cm$
$\sin A=\frac{B C}{A C}=\frac{12}{13}$
$\cos A=\frac{A B}{A C}=\frac{5}{13}$
$\frac{\cos A-\sin A}{\cos A+\sin A}$
$=\frac{\frac{5}{13}-\frac{12}{13}}{\frac{5}{13}+\frac{12}{13}}$
$=\frac{-\frac{7}{13}}{\frac{17}{13}}$
$=-\frac{7}{13} \times \frac{13}{17}$
$=-\frac{7}{17} .$
$ \therefore A C^2=A B^2+B C^2$
$\Rightarrow A B^2$
$=A C^2-B C^2$
$=13^2-12^2$
$=169-144$
$=25$
$\Rightarrow A B=5 \ cm$
$\sin A=\frac{B C}{A C}=\frac{12}{13}$
$\cos A=\frac{A B}{A C}=\frac{5}{13}$
$\frac{\cos A-\sin A}{\cos A+\sin A}$
$=\frac{\frac{5}{13}-\frac{12}{13}}{\frac{5}{13}+\frac{12}{13}}$
$=\frac{-\frac{7}{13}}{\frac{17}{13}}$
$=-\frac{7}{13} \times \frac{13}{17}$
$=-\frac{7}{17} .$
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