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

Given, \( 3 cot A \ = \ 4 \ = \ \frac{4}{3} \ = \ \frac{B}{P} \ = \ \frac{AB}{BC} \)

Let AB = 4k and BC = 3k

Then by Pythagoras theorem,

\( AC \ = \ \sqrt{AB^2 + BC^2} \ \)
\( = \ \sqrt{(4k)^2 + (3k)^2} \ \)
\( = \ \sqrt{16k^2 + 9k^2} \ = \ \sqrt{25k^2} \ \)
\( = \ 5k \)



\( sinA \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{3k}{5k} \ = \ \frac{3}{5} \)

\( cosA \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{4k}{5k} \ = \ \frac{4}{5} \)

and, \( tanA \ = \ \frac{1}{cotA} \ = \ \frac{3}{4} \)

Now, L.H.S. = \( \frac{1- tan^2A}{1+tan^2A} \ = \ \frac{1- \frac{9}{16}}{1+ \frac{9}{16}} \ = \ \frac{16 - 9}{16+9} \ = \ \frac{7}{25} \)

R.H.S. = \( cos^2A \ - \ sin^2A \ \)
\( = \ ( \frac{4}{5})^2 \ - \ ( \frac{3}{5})^2 \ \)
\( = \ \frac{16}
{25} \ - \ \frac{9}{25} \ \)
\( = \ \frac{7}{25} \)

\(\therefore \) L.H.S = R.H.S.

\(\therefore \) \( \frac{1 - tan^2A}{1+tan^2A} \ = \ cos^2A \ - \ sin^2A \)