TITLE: Finding $\alpha$ such that $\tan\alpha = \frac{(1+\tan 1^{\circ})\cdot (1+\tan 2^{\circ})-2}{(1-\tan 1^{\circ})\cdot(1-\tan 2^{\circ})-2}$
QUESTION [7 upvotes]: I am preparing for my college entry test and I ran into this problem in my book:
$$\tan\alpha = \frac{(1+\tan 1^{\circ})\cdot (1+\tan 2^{\circ})-2}{(1-\tan 1^{\circ})\cdot(1-\tan 2^{\circ})-2}$$
I should find the angle $\alpha$. Can anyone help me please? I tried solving it, but I just can't get any solution. By the way, the correct answer is $\alpha = 42^{\circ}$
Thanks!

REPLY [2 votes]: $$\tan\alpha=\dfrac{(\tan1^\circ\tan2^\circ-1)+(\tan1^\circ+\tan2^\circ)}{(\tan1^\circ\tan2^\circ-1)-(\tan1^\circ+\tan2^\circ)}$$
Using Componendo and Dividendo,$$\dfrac{\tan\alpha-1}{\tan\alpha+1}=\dfrac{\tan1^\circ+\tan2^\circ}{\tan1^\circ\tan2^\circ-1}$$
$$\iff\tan(\alpha-45^\circ)=-\tan(1^\circ+2^\circ)=\tan(-3^\circ)$$
$$\implies\alpha-45^\circ=180^\circ n+(-3^\circ)$$ where $n$ is any integer