Question

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

Answer 1

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

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

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

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

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

$=RHS$

Hence proved