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If $\tan\alpha = {(1+\tan1°)(1+\tan2°)-2 \over (1-\tan1°)(1-\tan2°) - 2}$ and $\alpha \in (0°, 90°)$ then $\alpha$ is equal to?

This is task from my faculty entrance exam workbook. This is mostly high school level and I can only assume that I need to use addition trigonometry formulas or double angle, but I do not how, may someone give me some steps or hint how to do this?

EDIT: I haven't done Componendo and Dividendo ever, is there any other way to solve this equation?

GreatDuke
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    When you pose a question here, it is expected that you include any work you have done on the problem and indicate where you are stuck so that you receive responses that address the specific difficulties you are encountering. – N. F. Taussig Jun 16 '15 at 09:18
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    Thanks for advice, I will keep that in mind for future questions and edit this one! – GreatDuke Jun 16 '15 at 09:24
  • try expand first – chenbai Jun 16 '15 at 09:34

2 Answers2

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HINT:

$$\tan\alpha=\dfrac{\tan x\tan y-1+(\tan x+\tan y)}{\tan x\tan y-1-(\tan x+\tan y)}$$

Applying Componendo and dividendo ,

$$\dfrac{\tan\alpha-1}{\tan\alpha+1}=\dfrac{\tan x+\tan y}{\tan x\tan y-1}$$

Now $$\dfrac{\tan x+\tan y}{\tan x\tan y-1}=-\dfrac{\tan x+\tan y}{1-\tan x\tan y}=-\tan(x+y)$$

and $$\dfrac{\tan\alpha-1}{\tan\alpha+1}=\tan\left[\alpha-45^\circ\right]$$

Finally, $\tan(-A)=-\tan A$

lab bhattacharjee
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  • No, $\tan\left[\alpha+45^{\circ}\right]=\frac{\tan\alpha+1}{1-\tan\alpha}$. – user26486 Jun 16 '15 at 10:02
  • @user26486, Thanks for your observation – lab bhattacharjee Jun 16 '15 at 10:04
  • Componendo and dividendo was applied in the form $$\frac{(\tan\alpha+1)+(\tan\alpha-1)}{(\tan\alpha+1)-(\tan\alpha-1)} =\frac{(\tan x\tan y-1)+(\tan x+\tan y)}{(\tan x\tan y-1)-(\tan x+\tan y)}$$ – user26486 Jun 16 '15 at 10:07
  • @user26486, Could you reach the destination ? – lab bhattacharjee Jun 16 '15 at 11:19
  • Componendo and dividendo says $$\frac{a}{b}=\frac{c}{d}\iff \frac{b+a}{b-a}=\frac{d+c}{d-c}$$

    Let $a=\tan\alpha-1, b=\tan\alpha+1,c=\tan x+\tan y,d=\tan x\tan y -1$.

     – user26486 Jun 16 '15 at 11:38
  • @user26486, Why don't you simplify to use $\tan(A\pm B)$ – lab bhattacharjee Jun 16 '15 at 11:44
  • You said you used componendo and dividendo (c&d), but you didn't show how you used it. If you insist on using c&d, then your statement $$\tan\alpha=\dfrac{\tan x\tan y-1+(\tan x+\tan y)}{\tan x\tan y-1-(\tan x+\tan y)}\implies \dfrac{\tan\alpha-1}{\tan\alpha+1}=\dfrac{\tan x+\tan y}{\tan x\tan y-1}$$

    can be proved by

    $$\frac{(\tan\alpha+1)+(\tan\alpha-1)}{(\tan\alpha+1)-(\tan\alpha-1)} =\frac{(\tan x\tan y-1)+(\tan x+\tan y)}{(\tan x\tan y-1)-(\tan x+\tan y)}\implies \dfrac{\tan\alpha-1}{\tan\alpha+1}=\dfrac{\tan x+\tan y}{\tan x\tan y-1}$$

     – user26486 Jun 16 '15 at 11:50
  • @user26486, As $\tan45^\circ=1,$ $$\dfrac{\tan\alpha-\tan45^\circ}{1+\tan\alpha\tan45^\circ}=\tan\left(\alpha-45^\circ\right)$$ right? Finally if $\tan x=\tan A,$ what can we conclude? – lab bhattacharjee Jun 16 '15 at 16:45
  • You said you used c&d, but you didn't give an explanation. You probably didn't use c&d because you don't even understand what I'm saying. – user26486 Jun 16 '15 at 16:54
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Given that $$\tan\alpha=\frac{(1+\tan 1^o)(1+\tan 2^o)-2}{(1-\tan 1^o)(1-\tan 2^o)-2}$$ $$=\frac{\left(1+\frac{\sin 1^o}{\cos 1^o}\right)\left(1+\frac{\sin 2^o}{\cos 2^o}\right)-2}{\left(1-\frac{\sin 1^o}{\cos 1^o}\right)\left(1-\frac{\sin 2^o}{\cos 2^o}\right)-2}$$ $$=\frac{(\sin 1^o+\cos 1^o)(\sin 2^o+\cos 2^o)-2\cos 1^o\cos 2^o}{(\sin 1^o-\cos 1^o)(\sin 2^o-\cos 2^o)-2\cos 1^o\cos 2^o}$$ $$=\frac{\sin 2^o\sin 1^o+\sin 2^o\cos 1^o+\cos 2^o\sin 1^o+\cos 2^o \cos 1^o-2\cos 2^o\cos 1^o}{\sin 2^o\sin 1^o-\sin 2^o\cos 1^o-\cos 2^o\sin 1^o+\cos 2^o \cos 1^o-2\cos 2^o\cos 1^o}$$ $$=\frac{(\sin 2^o\cos 1^o+\cos 2^o\sin 1^o)-(\cos 2^o\cos 1^o-\sin 2^o\sin 1^o)}{-(\sin 2^o\cos 1^o+\cos 2^o\sin 1^o)-(\cos 2^o\cos 1^o-\sin 2^o\sin 1^o)}$$ $$=\frac{\sin (2^o+1^o)-\cos (2^o+1^o)}{-\sin (2^o+1^o)-\cos (2^o+1^o)}$$ $$=\frac{\cos 3^o-\sin 3^o}{\cos 3^o+\sin 3^o}$$ $$=\frac{1-\tan 3^o}{1+\tan 3^o}$$ $$=\frac{\tan 45^o-\tan 3^o}{1+\tan 45^o\tan 3^o}=\tan(45^o-3^o)=\tan 42^o$$ $$\implies \tan \alpha=\tan 42^o $$$$\implies \color{blue}{\alpha=42^o} \quad (\forall \space 0<\alpha<90^o)$$

Harish Chandra Rajpoot
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