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Hint: - Here we go through by letting $\tan 3x$as $\tan (2x + x)$ . Because in question it is in terms of 2x and x. By applying this we can easily prove our question.

Let us take, $\tan 3x = \tan (2x + x)$

Now we can apply the formula of $\tan ({\text{A + B)}}$ i.e. $\tan ({\text{A + B) = }}\frac{{\tan {\text{A +

}}\tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}}$

Now we can write, $\tan 3x = \tan (2x + x) = \frac{{\tan 2x + \tan x}}{{1 - \tan 2x\tan x}}$

Now we cross multiply it to get,

$

\Rightarrow \tan 3x(1 - \tan 2x\tan x) = \tan 2x + \tan x \\

\Rightarrow \tan 3x - \tan 3x\tan 2x\tan x = \tan 2x + \tan x \\

$

Now by rearranging the above equation we get

$\tan 3x - \tan 2x - \tan x = \tan 3x\tan 2x\tan x$ Hence, proved.

Note: - Whenever we face such a type of question the key concept for solving the question is that we

always try to make the bigger angle in sum of two smaller angles that are given in a question to apply

the formula to prove the question.

Let us take, $\tan 3x = \tan (2x + x)$

Now we can apply the formula of $\tan ({\text{A + B)}}$ i.e. $\tan ({\text{A + B) = }}\frac{{\tan {\text{A +

}}\tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}}$

Now we can write, $\tan 3x = \tan (2x + x) = \frac{{\tan 2x + \tan x}}{{1 - \tan 2x\tan x}}$

Now we cross multiply it to get,

$

\Rightarrow \tan 3x(1 - \tan 2x\tan x) = \tan 2x + \tan x \\

\Rightarrow \tan 3x - \tan 3x\tan 2x\tan x = \tan 2x + \tan x \\

$

Now by rearranging the above equation we get

$\tan 3x - \tan 2x - \tan x = \tan 3x\tan 2x\tan x$ Hence, proved.

Note: - Whenever we face such a type of question the key concept for solving the question is that we

always try to make the bigger angle in sum of two smaller angles that are given in a question to apply

the formula to prove the question.