Prove that $\tan(6^\circ)\tan(42^\circ) = \tan(12^\circ) \tan(24^\circ)$.
I don't know how to approach this problem. One approach might be to note that $42-6= 24+12$, and then apply the identities for $\tan(A+B)$ and $\tan(A-B)$, but it just makes it more complex:
$$\frac{\tan(42^{\circ}) – \tan(6^{\circ})}{1+\tan(42^{\circ})\tan(6^{\circ})}= \frac{\tan(24^{\circ}) + \tan(12^{\circ})}{1 – \tan(24^{\circ})\tan(12^{\circ})}.$$
Can anyone here please provide a hint?
Best Answer
We can prove $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$ (Proof below)
Putting $x=6^\circ,$ $$\tan 6^\circ\tan 66^\circ\tan 54^\circ=\tan18^\circ$$
Putting $x=18^\circ,$ $$\tan 18^\circ\tan42^\circ\tan78^\circ=\tan54^\circ$$
On multiplication, $$\tan 6^\circ\tan 66^\circ\tan42^\circ\tan78^\circ=1$$
$$\tan 6^\circ\tan42^\circ=\cot66^\circ\cot78^\circ=\tan(90^\circ-66^\circ)\tan(90^\circ-78^\circ)=\tan24^\circ\tan12^\circ$$
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Proof $1:$ $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan x\cdot\frac{\tan60^\circ-\tan x}{1+\tan60^\circ\tan x}\cdot\frac{\tan60^\circ+\tan x}{1-\tan60^\circ\tan x}$$
$$=\tan x\cdot \frac{\sqrt3-\tan x}{1+\sqrt3\tan x}\cdot \frac{\sqrt3+\tan x}{1-\sqrt3\tan x}$$
$$=\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3x$$
Proof $2:$
If $\tan3x=\tan 3y$
$$\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3y$$
$$\implies \tan^3x-3\tan3y\tan^2x-3\tan x+\tan3y=0$$
If $\tan x_1,\tan x_2,\tan x_3$ are the three roots of above cubic equation,
$\implies \tan x_1\cdot\tan x_2\cdot\tan x_3=-\tan3y$
Again, as $\tan3x=\tan 3y$
$3x=180^\circ n+3y$ where $n$ is any integer $\implies x=60^\circ n+y$
So, the in-congruent values of $x$ are $y,60^\circ +y,60^\circ\cdot2+y$
So, the corresponding roots are $\tan y,\tan(60^\circ +y), \tan(60^\circ\cdot2+y)=\tan\{180^\circ-(60^\circ-y)\}=-\tan(60^\circ-y)$
$\implies -\tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=-\tan3y$
$\implies \tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=\tan3y$
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