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Hey guys! I'm 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 $\alpha = 42^{\circ}$


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Have you tried multiplying out the whole mess and exploiting, say, Pythagorean identities? – J. M. May 8 '11 at 16:12
Yeah. I still get some messy stuff. I also tried representing $\tan 2^{\circ}$ as $$\tan(2*1^{\circ}) = \frac {2\tan 1^{\circ}} {1-tan^2 1^{\circ}}$$ – Radiant May 8 '11 at 16:25
You could replace 1° by $x$ and 2° by $y$ and try to show that this expression of $\tan\alpha$ implies that $\alpha$ is $45°-x-y$. So... your first task could be to recall the formula for $\tan(45°-z)$. – Did May 8 '11 at 16:39
up vote 6 down vote accepted

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

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 $y=x$

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


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


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This is great! I can't thank you enough! – Radiant May 8 '11 at 17:30

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