Question

Show that: tan3A – tan2A – tanA = tan3A tan2A tanA.

Answer 1

- Hint: The formula for sum of two tangent terms, \[\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}\], can be used to evaluate tan3A and then simplified to get the desired result.

Complete step-by-step solution -

We will find the formula for the tangent of sum of angles from the formula for sine of sum of two angles and the formula for the cosine of sum of two angles.
The tangent of an angle can be written as the ratio of sine of the angle to the cosine of the angle.
\[\tan x = \dfrac{{\sin x}}{{\cos x}}.................(1)\]
The tangent of the sum of two angles can also be expressed in the same way as equation (1).
\[\tan (x + y) = \dfrac{{\sin (x + y)}}{{\cos (x + y)}}.........(2)\]
We know the formula for the sine of sum of two angles, that is, as follows:
\[\sin (x + y) = \sin x\cos y + \sin y\cos x..........(3)\]
We know the formula for the cosine of sum of two angles, that is, as follows:
\[\cos (x + y) = \cos x\cos y - \sin x\sin y..........(4)\]
Substituting equation (3) and equation (4) in equation (2), we have:
\[\tan (x + y) = \dfrac{{\sin x\cos y + \sin y\cos x}}{{\cos x\cos y - \sin x\sin y}}\]
Dividing numerator and denominator by cosx.cosy, we get:
\[\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}..........(5)\]
We now use formula (5) to evaluate tan3A as a tangent of sum of A and 2A, that is, tan(A+2A).
\[\tan (A + 2A) = \dfrac{{\tan A + \tan 2A}}{{1 - \tan A\tan 2A}}\]
Cross multiplying, we get:
\[\tan (3A)(1 - \tan A\tan 2A) = \tan A + \tan 2A\]
Multiplying inside the bracket, we get:
\[\tan 3A - \tan A\tan 2A\tan 3A = \tan A + \tan 2A\]
Now, taking the term tanA tan2A tan3A to right hand side and shifting the terms tanA and tan2A to the left-hand side, we have:
\[\tan 3A - \tan 2A - \tan A = \tan A\tan 2A\tan 3A\]
Hence, we showed that \[\tan 3A - \tan 2A - \tan A = \tan A\tan 2A\tan 3A\].

Note: You can also solve by considering the left-hand side of the equation and then simplifying it to express the right-hand side of the equation, which is more of a direct way to approach this question.