Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
3  
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
add comment

1 Answer

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 $y=x$

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

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

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

share|improve this answer
    
This is great! I can't thank you enough! –  Radiant May 8 '11 at 17:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.