Question

Find the value of tan 3A - tan2A - tanA is equal to ________

• A. tan 3A tan2A tanA
• B. -tan 3A tan 2A tanA
• C. tan A tan2A - tan2A + tan3A - tan3AtanA
• D. tan 3A + tan2A

Answer 1

The correct option is A

tan 3A tan2A tanA

We observe that the given expression contains the terms of tan3A,tan2A & tanA.

We can substitute the value of tan3A & tan2A.This is one of the ways.

Let's substitute the values and analyze what happens

tan3A - tan2A - tanA

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

= $\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$ - $\left(\frac{2tanA+tanA-ta{n}^{3}A}{1-ta{n}^{2}A}\right)$

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

= (3tanA - tan^3A) $\left[\frac{(3tanA-ta{n}^{3}A)\left(2ta{n}^{2}A\right)}{(1-3ta{n}^{2}A)(1-ta{n}^{2}A)}\right]$

We didn't see any pattern.If we want to proceed from this method, we have to substitute the value of tan3A & tan2A in each option and then check it.But this is not the effective method to solve these types of problem.

3A can be written as 2A + A

3A = 2A + A

tan3A = tan(2A + A)

we know the compound angle fromula for tan(A+B)

tan3A = $\frac{tan2A+tanA}{1-tan2A.tanA}$

Cross multiplying gives

tan3A - tan3A tan2A tanA = tan2A + tanA

tan3A - tan3A - tanA = tan3A tan2A tanA

Option A is correct.

This can save a lot of time.