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Worked Examples · Example 14

Q.Show that tan⁡3x tan⁡2x tan⁡x=tan⁡3x−tan⁡2x−tan⁡x\tan 3x\, \tan 2x\, \tan x = \tan 3x - \tan 2x - \tan x.

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Use the tangent addition formula tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} with A=2xA = 2x and B=xB = x to express tan⁡3x\tan 3x in terms of tan⁡2x\tan 2x and tan⁡x\tan x, then rearrange to isolate the product on one side.

The heart of this identity lies in recognizing that 3x=2x+x3x = 2x + x. When we apply the tangent addition formula to this decomposition, we create a relationship between tan⁡3x\tan 3x, tan⁡2x\tan 2x, and tan⁡x\tan x that naturally produces both their product and their sum. The key is algebraic manipulation: cross-multiplying and collecting terms strategically.

Why this works: The tangent addition formula connects the tangent of a sum to the tangents of individual angles through both addition and multiplication. Since our target identity mixes products and differences of tangents, the addition formula is the natural bridge.

  1. Start with the tangent addition formula

    Since 3x=2x+x3x = 2x + x, we write:

tan⁡3x=tan⁡(2x+x)=tan⁡2x+tan⁡x1−tan⁡2xtan⁡x\tan 3x = \tan(2x + x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}

  1. Cross-multiply to clear the denominator

    Multiply both sides by (1−tan⁡2xtan⁡x)(1 - \tan 2x \tan x):

tan⁡3x(1−tan⁡2xtan⁡x)=tan⁡2x+tan⁡x\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x

  1. Expand the left side

    Distribute tan⁡3x\tan 3x:

tan⁡3x−tan⁡3xtan⁡2xtan⁡x=tan⁡2x+tan⁡x\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x

  1. Rearrange to isolate the product term

    Move all single-tangent terms to the right side:

−tan⁡3xtan⁡2xtan⁡x=tan⁡2x+tan⁡x−tan⁡3x-\tan 3x \tan 2x \tan x = \tan 2x + \tan x - \tan 3x

  1. Multiply both sides by −1-1

    This gives us: …

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