Question

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

Answer 1

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