Question

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

Answer 1

$\frac{\sec \theta +\tan \theta -1}{\tan \theta -\sec \theta +1}$

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

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

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

$\Rightarrow \sec \theta +\tan \theta$

$\Rightarrow \frac{1}{\cos \theta }+\frac{\sin \theta }{\tan \theta }$

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

$\Rightarrow \frac{1+\sin \theta }{\cos \theta }\times \frac{1-\sin \theta }{1-\sin \theta }$

$\Rightarrow \frac{1-{\sin }^2\theta }{\cos \theta (1-\sin \theta )}$

$\Rightarrow \frac{{\cos }^2\theta }{\cos \theta (1-\sin \theta )}$

$\Rightarrow \frac{\cos \theta }{1-\sin \theta }$ (Hence proved)