Prove the following trigonometric identities.

In the given question, we need to prove $\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\cot ^{2} A}{1+\cot ^{2} A}=1$.

Here, we will first solve the LHS.

Now, using $\tan \theta=\frac{\sin \theta}{\cos \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$, we get

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

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

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

On further solving by taking the reciprocal of the denominator, we get,

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

$=\sin ^{2} A+\cos ^{2} A \quad\left(\right.$ Using $\left.\sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$=1$

Hence proved.