Question

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

Answer 1

R.H.S$=\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$

multiply numerator and denominator by $cos\theta$

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

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

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

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

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

$=\frac{1+\sin \theta }{cos\theta }=L.H.S$