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

Let we have a triangle, 

$Cot A=\frac{4}{3}$

Since $Cot \left(A\right) =\frac{Adjacent side}{Opposite side}=\frac{AB}{BC}=\frac{4}{3}$

Let AB be 4k and BC will be 3k

where k is a positive real number.

Using Pythagoras theorem, we get

$A{C}^2=A{B}^2 + B{C}^2$

${\left(AC\right)}^2={\left(4k\right)}^2 + {\left(3k\right)}^2$

$A{C}^2=16k^2+9k^2$

$A{C}^2=25k^2$

Thus, $AC =5k$

$Sin \left(A\right) =\frac{Opposite side}{Hypotenuse}= \frac{BC}{AC}=\frac{3k}{5k}=\frac{3}{5}$

$Cos \left(A\right) =\frac{Adjacent side}{Hypotenuse}= \frac{AB}{AC}= \frac{4k}{5k}=\frac{4}{5}$

$\tan \left(A\right) =\frac{Opposite side}{Adjacent side}=\frac{BC}{AB}=\frac{3k}{4k}=\frac{3}{4}$

So, $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}=\frac{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}{1+{\left(\frac{3}{4}\right)}^2}=\frac{1-{\displaystyle \frac{9}{16}}}{1+\frac{9}{16}}=\frac{16-9}{16+9}=\frac{7}{25}$

${\cos }^{2}A-{\sin }^{2}A={\left(\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$

Since both results are equal, the given statement is true.