Question

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

Answer 1

$\frac{\cos \theta -\frac{1}{\tan \theta }}{2\cot \theta }$

We know that,

$\cot \theta =\frac{1}{\tan \theta }$

$\frac{\cos \theta -\frac{1}{\tan \theta }}{2\cot \theta }$

$=\frac{\cos \theta -\cot \theta }{2\cot \theta }$

$=\frac{\cos \theta -\frac{\cos \theta }{\sin \theta }}{2\left(\frac{\cos \theta }{\sin \theta }\right)}$

$=\frac{\cos \theta }{\cos \theta }\times \frac{1-\frac{1}{\sin \theta }}{\frac{2}{\sin \theta }}$

$=\frac{1-\frac{1}{\sin \theta }}{\frac{2}{\sin \theta }}$

$=\frac{\frac{\sin \theta -1}{\sin \theta }}{\frac{2}{\sin \theta }}$

$=\frac{\sin \theta -1}{2}$

Given,

$\sin \theta =\frac{3}{5}$

$=\frac{\frac{3}{5}-1}{2}$

$=\frac{3-5}{10}$

$=\frac{-2}{10}=\frac{-1}{5}$