Question

Prove that $\frac{{\cos }^2\theta }{\sin \theta }-cosec\theta +\sin \theta =0$.

Answer 1

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

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

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

$=\frac{1-1}{\sin \theta }$

$=0$

$=RHS$

Hence proved