Prove the following trigonometric identities.

We need to prove $\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}=2 \sec ^{2} A$

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

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

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

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

So,

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

$=\frac{2 \operatorname{cosec}^{2} A}{\cot ^{2} A}$

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

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

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

$=2 \sec ^{2} A$

Hence proved.