If $ \sec \theta=\frac{5}{4} $, find the value of $ \frac{\sin \theta-2 \cos \theta}{\tan \theta-\cot \theta} $

Given:

$sec\ \theta = \frac{5}{4}$.

To do:

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

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $sec\ \theta = sec\ A=\frac{5}{4}$.

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,

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

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

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

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

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

Here,

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

$\Rightarrow (5)^2=(4)^2+BC^2$

$\Rightarrow BC^2=25-16$

$\Rightarrow BC=\sqrt{9}=3$

Therefore,

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

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

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

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

This implies,

Let us consider LHS,

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

$=\frac{\frac{3-8}{5}}{\frac{3( 3) -4( 4)}{12}}$

$=\frac{\frac{-5}{5}}{\frac{9-16}{12}}$

$=\frac{-1}{\frac{-7}{12}}$

$=\frac{12}{7}$

The value of \( \frac{\sin \theta-2 \cos \theta}{\tan \theta-\cot \theta} \) is \( \frac{12}{7} \).

Tutorialspoint

Updated on: 10-Oct-2022

913 Views

Kickstart Your Career

Get certified by completing the course

Get Started