Question

Prove that $\cot \theta -\tan \theta =\frac{2{\cos }^2\theta -1}{\sin \theta \cos \theta }$.

Answer 1

LHS$=\cot \theta -\tan \theta$

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

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

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

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

$=RHS$

Hence proved