Question

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

Answer 1

$L.H.Q.=\frac{\tan \theta -\cot \theta }{(\sin \theta \cdot \cos \theta )}$

$=\frac{\left(\frac{\sin \theta }{\cos \theta }\right)-\left(\frac{\cos \theta }{\sin \theta }\right)}{(\sin \theta \cdot \cos \theta )}$

$=\frac{\left(\frac{{\sin }^2\theta -{\cos }^2\theta }{(\sin \theta \cdot \cos \theta )}\right)}{(\sin \theta \cdot \cos \theta )}$

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

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

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

$={\sec }^2\theta -cose{c}^2\theta$

$=(1+{\tan }^2\theta )-(1+{\cot }^2\theta )$

$=1+{\tan }^2\theta -1-{\cot }^2\theta$

$={\tan }^2\theta -{\cot }^2\theta =RHQ$.