Question

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

Answer 1

$(1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )=(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta })(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })=\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)\left(\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right)=\frac{{(\sin \theta +\cos \theta )}^2-1}{\sin \theta \cos \theta }=\frac{\sin {}^2\theta +\cos {}^2\theta +2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }=\frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }=\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }=2$