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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.

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