Prove the following trigonometric identities.

We have to prove $\frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}=\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}=\tan ^{2} \theta$

Consider the expression

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

$=\frac{1+\tan ^{2} \theta}{\frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}}$

$=\tan ^{2} \theta \frac{1+\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$=\tan ^{2} \theta$

Again, we have

$\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}=\left(\frac{1-\tan \theta}{1-\frac{1}{\tan \theta}}\right)^{2}$

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

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

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

$=\tan ^{2} \theta(-1)^{2}$

$=\tan ^{2} \theta$