Question

Prove that :

$(1+\cot \theta +\tan \theta )(\sin \theta -\cos \theta )=\sin \theta \tan \theta -\cot \theta \cos \theta$

Answer 1

LHS:

$(1+\cot \theta +\tan \theta )(\sin \theta -\cos \theta )$

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

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

$=\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta }$

$=\frac{{\sin }^3\theta }{\sin \theta \cos \theta }-\frac{{\cos }^3\theta }{\sin \theta \cos \theta }$

$=\frac{\sin \theta .\sin \theta }{\cos \theta }-\frac{\cos \theta .\cos \theta }{\sin \theta }$

$=\sin \theta \tan \theta -\cot \theta \cos \theta$ = RHS