How to prove the following trigonometric equation?

Question

How to prove that $\tan5x.\tan3x.\tan 2x = \tan5x - \tan3x - \tan2x$

Answer 1

Since $3x+2x=5x$

Take $\tan$ on both sides $$\tan(3x+2x)=\tan5x$$ Use the trig formula $\tan(A+B)$ and we get $$\dfrac{\tan3x+\tan2x}{1-\tan3x\tan2x}=\tan5x$$ $$\tan3x+\tan2x=\tan5x(1-\tan3x\cdot\tan2x)$$ $$\tan3x+\tan2x=\tan5x-\tan2x\cdot\tan3x\cdot\tan5x$$ Therefore, $$\tan5x\cdot\tan3x\cdot\tan 2x = \tan5x - \tan3x - \tan 2x$$

Answer 2

$\tan(A+B+C)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{\cdots}$

If $A+B+C=n\pi,\tan(A+B+C)=?$

If all the tangent ratios are finite,

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

Here $A,B,C$ can be chosen

from $\{5x,-3x,-2x\}$

or from $\{-5x,3x,2x\}$

Use $\tan(-y)=-\tan y$