Find $a$, when $\tan a$ is given in terms of $\tan1^{\circ}$ and $\tan2^{\circ}$. If $\tan\alpha = {(1+\tan1°)(1+\tan2°)-2 \over (1-\tan1°)(1-\tan2°) - 2}$ and $\alpha \in (0°, 90°)$ then $\alpha$ is equal to?
This is task from my faculty entrance exam workbook. This is mostly high school level and I can only assume that I need to use addition trigonometry formulas or double angle, but I do not how, may someone give me some steps or hint how to do this?
EDIT: I haven't done Componendo and Dividendo ever, is there any other way to solve this equation?
 A: HINT:
$$\tan\alpha=\dfrac{\tan x\tan y-1+(\tan x+\tan y)}{\tan x\tan y-1-(\tan x+\tan y)}$$
Applying Componendo and dividendo ,
$$\dfrac{\tan\alpha-1}{\tan\alpha+1}=\dfrac{\tan x+\tan y}{\tan x\tan y-1}$$
Now $$\dfrac{\tan x+\tan y}{\tan x\tan y-1}=-\dfrac{\tan x+\tan y}{1-\tan x\tan y}=-\tan(x+y)$$
and $$\dfrac{\tan\alpha-1}{\tan\alpha+1}=\tan\left[\alpha-45^\circ\right]$$
Finally, $\tan(-A)=-\tan A$
A: Given that $$\tan\alpha=\frac{(1+\tan 1^o)(1+\tan 2^o)-2}{(1-\tan 1^o)(1-\tan 2^o)-2}$$ $$=\frac{\left(1+\frac{\sin 1^o}{\cos 1^o}\right)\left(1+\frac{\sin 2^o}{\cos 2^o}\right)-2}{\left(1-\frac{\sin 1^o}{\cos 1^o}\right)\left(1-\frac{\sin 2^o}{\cos 2^o}\right)-2}$$ $$=\frac{(\sin 1^o+\cos 1^o)(\sin 2^o+\cos 2^o)-2\cos 1^o\cos 2^o}{(\sin 1^o-\cos 1^o)(\sin 2^o-\cos 2^o)-2\cos 1^o\cos 2^o}$$
$$=\frac{\sin 2^o\sin 1^o+\sin 2^o\cos 1^o+\cos 2^o\sin 1^o+\cos 2^o \cos 1^o-2\cos 2^o\cos 1^o}{\sin 2^o\sin 1^o-\sin 2^o\cos 1^o-\cos 2^o\sin 1^o+\cos 2^o \cos 1^o-2\cos 2^o\cos 1^o}$$
$$=\frac{(\sin 2^o\cos 1^o+\cos 2^o\sin 1^o)-(\cos 2^o\cos 1^o-\sin 2^o\sin 1^o)}{-(\sin 2^o\cos 1^o+\cos 2^o\sin 1^o)-(\cos 2^o\cos 1^o-\sin 2^o\sin 1^o)}$$ $$=\frac{\sin (2^o+1^o)-\cos (2^o+1^o)}{-\sin (2^o+1^o)-\cos (2^o+1^o)}$$ $$=\frac{\cos 3^o-\sin 3^o}{\cos 3^o+\sin 3^o}$$ $$=\frac{1-\tan 3^o}{1+\tan 3^o}$$ $$=\frac{\tan 45^o-\tan 3^o}{1+\tan 45^o\tan 3^o}=\tan(45^o-3^o)=\tan 42^o$$ $$\implies \tan \alpha=\tan 42^o $$$$\implies \color{blue}{\alpha=42^o} \quad (\forall \space 0<\alpha<90^o)$$
