Prove the following trigonometric identities.

We have to prove

$(\operatorname{cosec} \theta-\sec \theta)(\cot \theta-\tan \theta)=(\operatorname{cosec} \theta+\sec \theta)(\sec \theta \operatorname{cosec} \theta-2)$

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

Consider the LHS.

$(\operatorname{cosec} \theta-\sec \theta)(\cot \theta-\tan \theta)=\left(\frac{1}{\sin \theta}-\frac{1}{\cos \theta}\right)\left(\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}\right)$

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

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

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

Now, consider the RHS.

$(\operatorname{cosec} \theta+\sec \theta)(\sec \theta \operatorname{cosec} \theta-2)=\left(\frac{1}{\sin \theta}+\frac{1}{\cos \theta}\right)\left(\frac{1}{\cos \theta} \frac{1}{\sin \theta}-2\right)$

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

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

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

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence proved.