Prove the following trigonometric identities.

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

Using $\cot \theta=\frac{\cos \theta}{\sin \theta}$ and $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$, we get

$1+\frac{\cot ^{2} \theta}{1+\operatorname{cosec} \theta}=\frac{1+\operatorname{cosec} \theta+\cot ^{2} \theta}{1+\operatorname{cosec} \theta}$

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

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

Further, using the property $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

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

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

$=\frac{1}{\sin \theta}$

$=\operatorname{cosec} \theta$

Hence proved.