Prove that

Solution:

Here we have to prove that $\frac{\tan \theta-\cot \theta}{\sin \theta \cos \theta}=\tan ^{2} \theta-\cot ^{2} \theta$

First we take LHS and use the identities

$\tan \theta=\frac{\sin \theta}{\cos \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$

$L H S=\frac{\tan \theta-\cot \theta}{\sin \theta \cos \theta}$

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

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

Now we take RHS

$R H S=\tan ^{2} \theta-\cot ^{2} \theta$

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

$=\frac{\sin ^{4} \theta-\cos ^{4} \theta}{\sin ^{2} \theta \cos ^{2} \theta}$

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

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

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

$=L H S$

Hence proved.