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

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!

Answer 1

$$\begin{align*} V&=\frac{\tan a\tan b+\tan a+\tan b-1}{\tan a\tan b-\tan a-\tan b-1}\\ &=\frac{\sin a \cos b + \cos a \sin b + \sin a\sin b - \cos a\cos b}{\sin a\sin b-\cos a\cos b-\sin a\cos b-\cos a\sin b}\\ &=\frac{\sin (a+b)-\cos(a+b)}{-\cos(a+b)-\sin(a+b)} \end{align*}$$

Now I will use: $$\sin x-\cos y = \sin x-\sin(90^\circ-y)=2\sin\frac{x+y-90^\circ}2\cos\frac{90^\circ+x-y}2,$$ which yields for $y=x$

$$\sin x-\cos x = -2\cos45^\circ\sin(45^\circ-x)$$

and

$$\cos x +\sin y = \sin(90^\circ-x)+\sin y = 2\cos\frac{90^\circ-x-y}2\sin\frac{90^\circ-x+y}2$$ which yields for $y=x$

$$\cos x+\sin y = 2\sin45^\circ\cos(45^\circ-x)$$

Plugging this into the above formula (for $x=a+b$) I get

$$V=\frac{-2\cos45^\circ\sin(45^\circ-a-b)}{-2\sin45^\circ\cos(45^\circ-a-b)}=\tan(45^\circ-a-b)$$

Answer 2

$$\tan\alpha=\dfrac{(\tan1^\circ\tan2^\circ-1)+(\tan1^\circ+\tan2^\circ)}{(\tan1^\circ\tan2^\circ-1)-(\tan1^\circ+\tan2^\circ)}$$

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