Prove the following identities,where the angles involved are acute angles for which the expressions are defined:
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

(A) First,consider the expression $\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$.
Using the identities $1+\tan ^{2} A = \sec ^{2} A$ and $1+\cot ^{2} A = \operatorname{cosec}^{2} A$,we get:
$\frac{\sec ^{2} A}{\operatorname{cosec}^{2} A} = \frac{1/\cos ^{2} A}{1/\sin ^{2} A} = \frac{\sin ^{2} A}{\cos ^{2} A} = \tan ^{2} A$.
Next,consider the expression $\left(\frac{1-\tan A}{1-\cot A}\right)^{2}$.
Substitute $\cot A = \frac{1}{\tan A}$:
$\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)^{2} = \left(\frac{1-\tan A}{\frac{\tan A - 1}{\tan A}}\right)^{2} = \left(\frac{(1-\tan A) \cdot \tan A}{-(1-\tan A)}\right)^{2} = (-\tan A)^{2} = \tan ^{2} A$.
Since both parts equal $\tan ^{2} A$,the identity is proved.