Question

If $\cos \theta$ = $\frac{3}{5}$, then evaluate $\frac{\sin \theta -\cot \theta }{2\tan \theta }$.

Answer 1

$\because$ $\cos \theta =\frac{3}{5}=\frac{BC}{AC}$
Let $BC=3k$ and $AC=5k$ where $k$ is any constant.
$\therefore$ From Baudhayana formula
$A{B}^2=A{C}^2-B{C}^2$
$=(5k{)}^2-(3k{)}^2$
$=25k^2-9k^2$
$=16k^2$
$\Rightarrow$ $AB=\pm 4k$
Since $\theta$ is the acute angle, therefore
$\therefore$ $\sin \theta =\frac{AB}{AC}=\frac{4k}{5k}=\frac{4}{5}$
$\tan \theta =\frac{AB}{BC}=\frac{4k}{3k}=\frac{4}{3}$
and $\cot \theta =\frac{3}{4}$
$\therefore$ The value of the expression $\frac{\sin \theta -\cot \theta }{2\tan \theta }$
$=\frac{\frac{4}{5}-\frac{3}{4}}{2\times \frac{4}{3}}=\frac{\frac{16-15}{20}}{\frac{8}{3}}$
$=\frac{1}{20}\times \frac{3}{8}=\frac{3}{160}$