Question

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

Answer 1

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

$\frac{(sec\theta -1)+tan\theta }{-(sec\theta -1)+tan\theta }\Rightarrow \frac{se{c}^2\theta +1-2sec\theta +ta{n}^2\theta +2sin\theta -2tan\theta }{ta{n}^2\theta -se{c}^2\theta -1+2sec\theta }$

$\Rightarrow \frac{1+ta{n}^2\theta +1-2sec\theta +ta{n}^2\theta +2sin\theta -tan\theta }{ta{n}^2\theta -1-ta{n}^2\theta -1+2sec\theta }$

$\Rightarrow \frac{2+2ta{n}^2\theta -2sec\theta -2tan\theta +2sin\theta }{-2+2sec\theta }$

$\Rightarrow \frac{2+2se{c}^2-2-2sec\theta -2tan\theta +2sin\theta }{-2+2sec\theta }$

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

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

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