Question

If $\sin \theta =\frac{3}{5}$ then show that $\frac{\cos \theta -\frac{1}{\tan \theta }}{2\cot \theta }=-\frac{1}{5}$

Answer 1

$\sin \theta =\frac{3}{5}$ $\sqrt{25-9}=\sqrt{16}=4$

From triangle $\cos \theta =\frac{4}{5}$

$\tan \theta =\frac{3}{4}\&\cot \theta =\frac{4}{3}$

$\frac{\cos \theta -\frac{1}{\tan \theta }}{2\cot \theta }$ substituting values in this expression

$\frac{\frac{4}{5}-\frac{4}{3}}{2\times \frac{4}{3}}=\frac{-\frac{8}{15}}{\frac{8}{3}}=-\frac{1}{5}$

Hence, proved