Question

prove that $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$ using the identity ${\sec }^2\theta =1+{\tan }^2\theta$.

Answer 1

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

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

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

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

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

$=\frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }=\frac{1}{\sec \theta -\tan \theta }=RHS$