If $ \cos \theta=\frac{5}{13} $, find the value of $ \frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta} $

Given:

\( \cos \theta=\frac{5}{13} \).

To do:

We have to find the value of \( \frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta} \).

Solution:  

Let, in a triangle $ABC$ right-angled at $B$ and $\ cos\ \theta = cos\ A=\frac{5}{13}$.

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,

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

$\Rightarrow (13)^2=(5)^2+BC^2$

$\Rightarrow BC^2=169-25$

$\Rightarrow BC=\sqrt{144}=12$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{12}{13}$

$tan\ \theta=\frac{BC}{AB}=\frac{12}{5}$

This implies,

$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta}=\frac{\left(\frac{12}{13}\right)^{2} -\left(\frac{5}{13}\right)^{2}}{2\left(\frac{12}{13}\right)\left(\frac{5}{13}\right)} \times \frac{1}{\left(\frac{12}{5}\right)^{2}}$

$=\frac{\frac{144-25}{169}}{\frac{120}{169}} \times \frac{25}{144}$

$=\frac{119}{120} \times \frac{25}{144}$

$=\frac{119\times 5}{24\times 144}$

$=\frac{595}{3456}$

The value of \( \frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta} \times \frac{1}{\tan ^{2} \theta} \) is \( \frac{595}{3456} \).