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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 $\alpha = 42^{\circ}$

Thanks!

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    If the problem is simply to evaluate that, why not do the obvious: use a calculator to find tan(1) and tan(2), evaluate the right hand side, then take the arctangent?2016-09-10

2 Answers 2

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

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

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

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    This is great! I can't thank you enough!2011-05-08
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$\tan\alpha=\dfrac{(\tan1^\circ\tan2^\circ-1)+(\tan1^\circ+\tan2^\circ)}{(\tan1^\circ\tan2^\circ-1)-(\tan1^\circ+\tan2^\circ)}$

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