Question

Prove: $\frac{\sec \theta +1-\tan \theta }{\sec \theta +1+\tan \theta }+\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=2\sec \theta$

Answer 1

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

$=(\sec \theta -\tan \theta )\frac{[1+\sec \theta +\tan \theta ]}{[1+\sec \theta +\tan \theta ]}=\sec \theta -\tan \theta =\frac{1-\sin \theta }{\cos \theta }...\left(1\right)$

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

$=\sec \theta +\tan \theta \frac{[1-\sec \theta +\tan \theta ]}{1-\sec \theta +\tan \theta }=\sec \theta +\tan \theta =\frac{1+\sin \theta }{\cos \theta }...\left(2\right)$

$\left(1\right)+\left(2\right)=\frac{2}{\cos \theta }=2\sec \theta$