tan3A is equal to

Expression : $$tan(3A)$$

= $$\frac{sin(3A)}{cos(3A)} = \frac{sin(2A + A)}{cos(2A + A)}$$

= $$\frac{sin(2A)cosA + cos(2A)sinA}{cos(2A)cosA - sin(2A)sinA}$$

Using, $$sin(2x) = 2sinxcosx$$ and $$cos(2x) = cos^2x-sin^2x$$

= $$\frac{2sinAcos^2A + sinA(cos^2A-sin^2A)}{cosA(cos^2A-sin^2A) - 2sin^2AcosA}$$

Taking $$sinA$$ common from numerator and $$cosA$$ from denominator,

= $$\frac{sinA}{cosA} \frac{2cos^2A + cos^2A - sin^2A}{cos^2A - sin^2A - 2sin^2A}$$

= $$tanA \frac{3cos^2A - sin^2A}{cos^2A - 3sin^2A}$$

Dividing both numerator and denominator by $$cos^2A$$, we get :

= $$tanA \frac{3 - tan^2A}{1 - 3tan^2A}$$

= $$\frac{3tanA - tan^3A}{1 - 3tan^2A}$$

=> Ans - (A)