The absolute of the expression$\frac{1+\sqrt3 \tan 1^o}{\sqrt3 -\tan 1^o} \cdot \frac{\tan 89^o}{\cot 3^o}$ is equal to...

Question

Question: The absolute of the expression$\frac{1+\sqrt3 \tan 1^o}{\sqrt3 -\tan 1^o} \cdot \frac{\tan 89^o}{\cot 3^o}$ is equal to...

What I have tried...

Converting all the tan in fraction to cot... $$\frac{1+\sqrt3 \cot 89^o}{\cot 30^o -\cot 89^o} \cdot \frac{\cot 1^o}{\cot 3^o}$$ $$\cot (30^o - 89^o) \cdot \frac{\cot 1^o}{\cot 3^o}$$

But I can't seem to get an absolute value of the expression. The steps that I have done here may be wrong. The answer to this question is 1. Please help. Thank you!

Answer 1

You can prove that $\frac{\tan 3x}{\tan x}=\tan(60^\circ-x)\tan(60^\circ+x)$, so $\frac{\tan 89^\circ}{\cot 3^\circ}= \frac{\tan 3^\circ}{\tan 1^\circ}= \tan 59^\circ\tan 61^\circ$. And $\frac{1+\sqrt 3\tan 1^\circ}{\sqrt 3-\tan 1^\circ}= \frac{1+\tan 60^\circ\tan 1^\circ}{\tan 60^\circ-\tan 1^\circ}= \frac{1}{\tan(60^\circ-1^\circ)}=\frac{1}{\tan 59^\circ}$. Thus the expression equals $\tan 61^\circ$.