Question

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

Answer 1

$LHS$

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

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

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

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

$=\frac{1+{\sin }^2\theta +2\sin \theta -{\cos }^2\theta }{{\sin }^2\theta +2\sin \theta \cos \theta +{\cos }^2\theta -1}$

$=\frac{2{\sin }^2\theta +2\sin \theta }{2\sin \theta \cos \theta }$

$[\because {\sin }^{2}A+{\cos }^{2}A=1]$

$=\frac{2\sin \theta (\sin \theta +1)}{2\sin \theta \cos \theta }$

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

$=\frac{\sin \theta +1}{\cos \theta }\times \frac{(-\sin \theta +1)}{(-\sin \theta +2)}$

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

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

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

$=\sec \theta -\tan \theta$

$=RHS$

So, $LHS=RHS$