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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 \( \frac{312}{25} \).