# If $\tan \theta=\frac{12}{13}$, find the value of $\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$

Given:

$tan\ \theta = \frac{12}{13}$.

To do:

We have to find the value of $\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$.

Solution:

Let, in a triangle $ABC$ right-angled at $B$ and $\ tan\ \theta = tan\ A=\frac{12}{13}$.

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,

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

$\Rightarrow AC^2=(13)^2+(12)^2$

$\Rightarrow AC^2=169+144$

$\Rightarrow AC=\sqrt{313}$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{12}{\sqrt{313}}$

$cos\ \theta=\frac{AB}{AC}=\frac{13}{\sqrt{313}}$

This implies,

$\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{2\times \left(\frac{12}{\sqrt{313}}\right) \times \left(\frac{13}{\sqrt{313}}\right)}{\left(\frac{13}{\sqrt{313}}\right)^{2} -\left(\frac{12}{\sqrt{313}}\right)^{2}}$

$=\frac{\frac{312}{313}}{\frac{169-144}{313}}$

$=\frac{312}{25}$

The value of $\frac{2 \sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$ is $\frac{312}{25}$.

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Updated on: 10-Oct-2022

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