Question

Prove the following:
$(1+\cot \theta -cosec\theta )(1+\tan \theta +\sec \theta )=2$

Answer 1

L.H.S $=(1+\cot \theta -cosec\theta )(1+\tan \theta +\sec \theta )$
$=(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta })(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })$
$=\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)\left(\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right)$
$=\left(\frac{(\sin \theta +\cos \theta )-1}{\sin \theta }\right)\left(\frac{(\sin \theta +\cos \theta )+1}{\cos \theta }\right)=\frac{\sin \theta +\cos \theta {)}^2-1}{\sin \theta \cos \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }=\frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }=\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }=2$