Question

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

Answer 1

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

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

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

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

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

$LHS=RHS$