If $3\ cot\ \theta = 2$, find the value of $\frac{4\ sin\ θ−3\ cos\ θ}{2\ sin\ θ+6\ cos\ θ}$.

Given:

$3\ cot\ \theta = 2$.

To do:

We have to find the value of $\frac{4\ sin\ θ−3\ cos\ θ}{2\ sin\ θ+6\ cos\ θ}$.

Solution:  

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

$3\ cot\ \theta = 2$

$cot\ \theta = cot\ A= \frac{2}{3}$

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,

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

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

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

Here,

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

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

$\Rightarrow AC^2=4+9$

$\Rightarrow AC=\sqrt{13}$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{3}{\sqrt{13}}$

$cos\ \theta=\frac{AB}{AC}=\frac{2}{\sqrt{13}}$

This implies,

 $\frac{4\ sin\ θ−3\ cos\ θ}{2\ sin\ θ+6\ cos\ θ}=\frac{4\left(\frac{3}{\sqrt{13}}\right) -3\left(\frac{2}{\sqrt{13}}\right)}{2\left(\frac{3}{\sqrt{13}}\right) +6\left(\frac{2}{\sqrt{13}}\right)}$

$=\frac{\frac{12-6}{\sqrt{13}}}{\frac{6+12}{\sqrt{13}}}$

$=\frac{6}{18}$

$=\frac{1}{3}$

The value of $\frac{4sin\ θ-3cos\ θ}{2sin\ θ+6cos\ θ}$ is $\frac{1}{3}$.

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Updated on: 10-Oct-2022

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