Question

Show that \[\tan 3x\tan 2x\tan x=\tan 3x-\tan 2x-\tan x\] .

Answer 1

Hint: When an equation is given in terms of sine, cosine, tangent, we must use any of the trigonometric identities to make the inequation solvable. There are inter relations between sine, cosine, tan, secant these are inter relations are called as identities. Whenever you can see conditions such that \[\theta \in R\] , that means inequality is true for all angles. So, directly think of identity which will make your work easy.
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] for all \[\theta \] , \[\tan \left( A+B
\right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
An equality with sine, cosine or tangent in them is called a trigonometric equation. These are solved by some inter relations known beforehand.
All the inter relations which relate sine, cosine, tangent, secant, cotangent are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.

Complete step-by-step answer:
Given equation in the question which we need to prove:

\[\tan 3x\tan 2x\tan x=\tan 3x-\tan 2x-\tan x\]

By general algebra, we can write 3x as \[2x+x\] ; we get

\[3x=2x+x\]

By applying tan on both sides of equation, we get:

\[\tan 3x=\tan \left( 2x+x \right)\]

By general knowledge of trigonometry, we know formula of \[\tan \left( A+B \right)\] :

\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]

By substituting this into equation above, we get \[\tan 3x\] as:

\[\tan 3x=\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}\]

By cross multiplying the terms in above, we get it as:

\[\tan 3x\left( 1-\tan x\tan 2x \right)=\tan 2x+\tan x\]

By simplifying the above equation, the LHS will turn to:

\[\tan 3x-\tan x\tan 2x\tan 3x=\tan x+\tan 2x\]

By rearranging the terms in equation, we get it as:

\[\tan 3x\tan 2x\tan x=\tan 3x-\tan 2x-\tan x\]

This is the required equation which is asked in the question.

Hence, we proved the required equation.

Note: By substituting formulas of \[\tan 3x,\tan 2x\] we can solve the given equation. This is an alternate method but it will be very long to solve.