If $3\ tan\ θ = 4$, find the value of $\frac{4\ cos\ θ – sin\ θ}{2\ cos\ θ+sin\ θ}$.

Given:

$3tan\ θ = 4$.

To do:

We have to find the value of $\frac{4\ cos\ θ – sin\ θ}{2\ cos\ θ+sin\ θ}$.

Solution:  

Let, in a triangle $ABC$ right-angled at $B$ and $\ tan\ \theta = tan\ A$.

$3\ tan\ \theta = 4$

$tan\ \theta = tan\ A=\frac{4}{3}$

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,

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

$\Rightarrow AC^2=(3)^2+(4)^2$

$\Rightarrow AC^2=9+16$

$\Rightarrow AC=\sqrt{25}=5$

Therefore,

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

$cos\ \theta=\frac{AB}{AC}=\frac{3}{5}$

$\frac{4\ cos\ θ – sin\ θ}{2\ cos\ θ+sin\ θ}=\frac{4\left(\frac{3}{5}\right) -\left(\frac{4}{5}\right)}{2\left(\frac{3}{5}\right) +\frac{4}{5}}$

$=\frac{\frac{12-4}{5}}{\frac{6+4}{5}}$

$=\frac{8}{10}$

$=\frac{4}{5}$

The value of $\frac{4\ cos\ θ-sin\ θ}{2\ cos\ θ+sin\ θ}$ is $\frac{4}{5}$.

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Updated on: 10-Oct-2022

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