If $ \cos \theta=\frac{3}{5} $, find the value of $ \frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta} $

Given:

\( \cos \theta=\frac{3}{5} \)

To do:

We have to find the value of \( \frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta} \).

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $cos\ \theta = cos\ A=\frac{3}{5}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow (5)^2=(3)^2+BC^2$

$\Rightarrow BC^2=25-9$

$\Rightarrow BC=\sqrt{16}=4$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{4}{5}$

$tan\ \theta=\frac{BC}{AB}=\frac{4}{3}$

$\Rightarrow \frac{1}{tan\ \theta}=\frac{1}{\frac{4}{3}}=\frac{3}{4}$

This implies,

$\frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta}=\frac{\left(\frac{4}{5}\right) -\left(\frac{3}{4}\right)}{2\left(\frac{4}{3}\right)}$

$=\frac{\frac{4( 4) -3( 5)}{5( 4)}}{\frac{8}{3}}$

$=\frac{\frac{16-15}{20}}{\frac{8}{3}}$

$=\frac{1}{20} \times \frac{3}{8}$

$=\frac{3}{160}$

The value of \( \frac{\sin \theta-\frac{1}{\tan \theta}}{2 \tan \theta} \) is \( \frac{3}{160} \).