Question

Prove that \[\tan 3x.\tan 2x.\tan x = \tan 3x - \tan 3x - \tan x\]?

Answer 1

Hint: Here we have a trigonometric problem. Since here we have tangent function we need to know the tangent formulas. We take \[\tan 3x\] and we split the angle as a sum of two numbers. Then here we use the sum identity of tangent to prove the given problem. That is \[\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x.\tan y}}\].

Complete step-by-step solution:
Now take
\[\tan 3x\].
We can write \[3x = 2x + x\], then
\[\tan \left( {3x} \right) = \tan \left( {2x + x} \right)\]
Now applying \[\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x.\tan y}}\] , we have
\[\tan \left( {3x} \right) = \dfrac{{\tan 2x + \tan x}}{{1 - \tan 2x.\tan x}}\].
Now cross multiplying we have,
\[\tan \left( {3x} \right)\left( {1 - \tan 2x.\tan x} \right) = \tan 2x + \tan x\]
Expanding the brackets we have,
\[\tan 3x - \tan 3x.\tan 2x.\tan x = \tan 2x + \tan x\]
Now rearranging the equation we have,
 \[\tan 3x - \tan 2x - \tan x = \tan 3x.\tan 2x.\tan x\]
Or
\[ \Rightarrow \tan 3x.\tan 2x.\tan x = \tan 3x - \tan 2x - \tan x\].
Hence proved.

Note: Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. We also know the difference identity for the tangent function. That is \[\tan (x - y) = \dfrac{{\tan x - \tan y}}{{1 + \tan x.\tan y}}\].