If $3 \cot A = 4$,check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.

(A) It is given that $3 \cot A = 4$.
Therefore,$\cot A = \frac{4}{3}$.
Consider a right triangle $ABC$,right-angled at point $B$.
$\cot A = \frac{\text{Side adjacent to } \angle A}{\text{Side opposite to } \angle A} = \frac{AB}{BC} = \frac{4}{3}$.
Let $AB = 4k$ and $BC = 3k$,where $k$ is a positive constant.
In $\triangle ABC$,by Pythagoras theorem:
$(AC)^2 = (AB)^2 + (BC)^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2$.
Thus,$AC = 5k$.
Now,$\cos A = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$.
$\sin A = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5}$.
$\tan A = \frac{BC}{AB} = \frac{3k}{4k} = \frac{3}{4}$.
$LHS$: $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.
$RHS$: $\cos^2 A - \sin^2 A = (4/5)^2 - (3/5)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.
Since $LHS$ = $RHS$,the given equation is true.