Question

Prove that:- $(1+\cot \theta -\cos ec\theta )(1+\tan \theta +sec\theta )$ = $2$

Answer 1

L.H.S

$(1+\cot \theta -\cos ec\theta )(1+\tan \theta +sec\theta )$

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

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

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

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

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

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

$=2$

R.H.S

Hence, proved.