# If 3 cot A = 4, check whether $$1 – tan^2A\over 1 + tan^2A$$ = $$cos^2A – sin^2A$$ or not.

## Solution :

We have,   3 cot A = 4    $$\implies$$  cot A = $$4\over 3$$ = $$AB\over BC$$

Let  AB = 4k, then BC = 3k

By Pythagoras Theorem,

$${AC}^2$$ = $${AB}^2$$ + $${BC}^2$$

$$\implies$$  $${AC}^2$$ = $$25k^2$$

$$\implies$$  AC = 5k

Thus,  tan A = $$BC\over AB$$ = $$3k\over 4k$$ = $$3\over 4$$

sin A = $$BC\over AC$$ = $$3k\over 5k$$ = $$3\over 5$$

cos A = $$AB\over AC$$ = $$4k\over 5k$$ = $$4\over 5$$

Now, $$1 – tan^2A\over 1 + tan^2A$$ = $$16 – 9\over 16 + 9$$ = $$7\over 25$$               …………(1)

and  $$cos^2A – sin^2A$$ = $$7\over 25$$                    ………….(2)

From (1) and (2), we have :

$$1 – tan^2A\over 1 + tan^2A$$ = $$cos^2A – sin^2A$$