Question

Prove that $\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{1+\sin \theta }{\cos \theta }$

Answer 1

To Prove:

$\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{1+\sin \theta }{\cos \theta }$

Solution:

L.H.S $=\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}$

We can write, ${\sec }^2\theta -{\tan }^2\theta =1$

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

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

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

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

$=\tan \theta +\sec \theta$

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

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

$=$ R.H.S

since L.H.S $=$ R.H.S

$\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{1+\sin \theta }{\cos \theta }$

Hence Proved.