Question

Show that:

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

Answer 1

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

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

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

$=\cot \theta -\tan \theta$.