If $ \tan \theta=\frac{12}{5} $, find the value of $ \frac{1+\sin \theta}{1-\sin \theta} $

Given:

\( \tan \theta=\frac{12}{5} \)

To do:

We have to evaluate \( \frac{1+\sin \theta}{1-\sin \theta} \).

Solution:  

We know that,

$\sec ^{2} \theta=1+\tan ^{2} \theta$

$\cos ^{2} \theta=\frac{1}{\sec ^{2} \theta}$

$=\frac{1}{1+\tan ^{2} \theta}$
$\Rightarrow \cos \theta=\frac{1}{\sqrt{1+\tan ^{2} \theta}}$

$=\frac{1}{\sqrt{1+(\frac{12}{5})^2}}$

$=\frac{1}{\sqrt{1+\frac{144}{25}}}$

$=\frac{1}{\sqrt{\frac{25+144}{25}}}$

$=\frac{1}{\sqrt{\frac{169}{25}}}$

$=\frac{1}{\frac{13}{5}}$

$=\frac{5}{13}$

$\sin \theta=\sqrt{1-\cos^2 \theta}$

$=\sqrt{1-(\frac{5}{13})^2}$

$=\sqrt{\frac{169-25}{169}}$

$=\sqrt{\frac{144}{169}}$

$=\frac{12}{13}$

This implies,

$\frac{1+\sin \theta}{1-\sin \theta}=\frac{1+\frac{12}{13}}{1-\frac{12}{13}}$

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

$=\frac{25}{1}$

$=25$

The value of \( \frac{1+ \sin \theta}{1- \sin \theta} \) is $25$. 

Tutorialspoint

Updated on: 10-Oct-2022

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