Prove the following identities:$ \frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta} $

To do:

We have to prove that \( \frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta} \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

$\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1+(\sec \theta-\tan \theta)}{1+\sec \theta+\tan \theta}$

$=\frac{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)+(\sec \theta-\tan \theta)}{1+\sec \theta+\tan \theta}$

$=\frac{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)+(\sec \theta-\tan \theta)}{1+\sec \theta+\tan \theta}$

$=\frac{(\sec \theta-\tan \theta)[\sec \theta+\tan \theta+1]}{1+\sec \theta+\tan \theta}$

$=\sec \theta-\tan \theta$

$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$

$=\frac{1-\sin \theta}{\cos \theta}$

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

94 Views

Kickstart Your Career

Get certified by completing the course

Get Started