Prove the following trigonometric identities.

In the given question, we need to prove $\frac{\cos \theta}{\operatorname{cosec} \theta+1}+\frac{\cos \theta}{\operatorname{cosec} \theta-1}=2 \tan \theta$

Using the identity $a^{2}-b^{2}=(a+b)(a-b)$, we get

$\frac{\cos \theta}{\operatorname{cosec} \theta+1}+\frac{\cos \theta}{\operatorname{cosec} \theta-1}=\frac{\cos \theta(\operatorname{cosec} \theta-1)+\cos \theta(\operatorname{cosec} \theta+1)}{\operatorname{cosec}^{2} \theta-1}$

$=\frac{\cos \theta(\operatorname{cosec} \theta-1+\operatorname{cosec} \theta+1)}{\operatorname{cosec}^{2} \theta-1}$

Further, using the property $1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta$, we get

$\frac{\cos \theta(\operatorname{cosec} \theta-1+\operatorname{cosec} \theta+1)}{\operatorname{cosec}^{2} \theta-1}=\frac{\cos \theta(2 \operatorname{cosec} \theta)}{\cot ^{2} \theta}$

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

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

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

$=2 \tan \theta$

Hence proved.