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

Given:

\( \sin \theta=\frac{3}{5} \).

To do:

We have to evaluate \( \frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta} \).

Solution:  

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

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,

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

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

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

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

Here,

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

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

$\Rightarrow AB^2=25-9$

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

Therefore,

$cos\ \theta=\frac{AB}{AC}=\frac{4}{5}$

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

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

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

This implies,

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

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

$=\frac{\frac{12-20}{15}}{\frac{8}{3}}$

$=\frac{-8}{15} \times \frac{3}{8}$

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

The value of \( \frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta} \) is \( \frac{-1}{5} \). 

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Updated on: 10-Oct-2022

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