<s>[INST] Question related to tan in a ratio in a triangle If in a triangle $\tan A:\tan B:\tan C = 1:2:3$ then, what are the ratio of the sides $a,b,c $?
 [/INST] $$
sinA = \frac{a}{2R}
$$
where R is the radius of the circumcircle. It'll be cancelled out later.
$$
cosA = \frac{b^2+c^2-a^2}{2bc}
$$
So,
$$
tanA = \frac{abc}{R(b^2+c^2-a^2)}
$$
And similarly,
$$
tanB = \frac{abc}{R(a^2+c^2-b^2)},tanC = \frac{abc}{R(a^2+b^2-c^2}
$$
Canceling out $\frac{abc}{R}$ from the ratio, we get,
$$
\frac{1}{b^2+c^2-a^2}:\frac{1}{a^2+c^2-b^2}:\frac{1}{a^2+b^2-c^2} = 1:2:3
$$
Equating each of these terms to k, 2k and 3k respectively (so that the ratio 1:2:3 is maintained), we get the following equations
$$
b^2+c^2-a^2 = \frac{1}{k}
$$
$$
a^2+c^2-b^2 = \frac{1}{2k}
$$
$$
a^2+b^2-c^2 = \frac{1}{3k}
$$
On solving these, we get $a=\frac{\surd{5}}{2\surd{3}k}$, $ b=\frac{\surd{2}}{\surd{3}k} $ and $c=\frac{\surd{3}}{2k}$
Therefore $a:b:c = \surd{5}:2\surd{2}:3$
</s>