Question

Prove that :
$(1+\tan \theta +\sec \theta )(1+\cot \theta -\text{cosec}\theta )=2$.

Answer 1

$(1+\tan \theta +\sec \theta )(1+\cot \theta -\text{cosec}\theta )$

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

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

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

$=\frac{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$ ........ [Since ${\sin }^2\theta +{\cos }^2\theta =1$]

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