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Question

Question: Show that: \(\tan 3x - \tan 2x - \tan x = \tan 3x\tan 2x\tan x\)...

Show that: tan⁡3x−tan⁡2x−tan⁡x=tan⁡3xtan⁡2xtan⁡x\tan 3x - \tan 2x - \tan x = \tan 3x\tan 2x\tan x

Explanation

Solution

Hint: - Here we go through by letting tan⁡3x\tan 3xas tan⁡(2x+x)\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)\tan 3x = \tan (2x + x)
Now we can apply the formula of tan⁡(A + B)\tan ({\text{A + B)}} i.e. tan⁡(A + B) = tan⁡A + tan⁡B1−tan⁡Atan⁡B\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)=tan⁡2x+tan⁡x1−tan⁡2xtan⁡x\tan 3x = \tan (2x + x) = \frac{{\tan 2x + \tan x}}{{1 - \tan 2x\tan x}}
Now we cross multiply it to get,

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