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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,
$sin\ \theta=\frac{BC}{AC}=\frac{4}{5}$
$cos\ \theta=\frac{AB}{AC}=\frac{3}{5}$
This implies,$\frac{4\ cos\ θ – sin\ θ}{2\ cos\ θ+sin\ θ}=\frac{4\left(\frac{3}{5}\right) -\left(\frac{4}{5}\right)}{2\left(\frac{3}{5}\right) +\frac{4}{5}}$
$=\frac{\frac{12-4}{5}}{\frac{6+4}{5}}$
$=\frac{8}{10}$
$=\frac{4}{5}$
The value of $\frac{4\ cos\ θ-sin\ θ}{2\ cos\ θ+sin\ θ}$ is $\frac{4}{5}$.