# If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$

If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$

If $\alpha +\beta = \dfrac{\pi}{4}$ then $\tan(\alpha +\beta) = \tan(\dfrac{\pi}{4}) = 1$

We also know that $\tan(\alpha +\beta) = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$

Then we can write $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$

I have tried rearranging $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$ but it has not been helpful.

I also thought if we let $\alpha = \beta$ then I could write $\tan(\alpha+ \alpha) = 1$ (does this also mean $\tan(2\alpha) = 1$?)

then: $\tan(\alpha + \alpha) = \dfrac{\tan\alpha + \tan\alpha}{1- \tan\alpha\tan\alpha}$

which gives: $1 = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$

Anyway these are my thoughts so far, any hints would be really appreciated.

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You can't say $\alpha = \beta$. You can always use $\beta = \frac{\pi}{4} - \alpha$, substitute that in begging equation, but there is probably faster way. – Cortizol Jul 18 '13 at 7:40
– lab bhattacharjee Jul 18 '13 at 8:11
Just one step more, namely $1-\tan{\alpha}\tan{\beta}=\tan{\alpha}+\tan{\beta}$! – Ali Jul 18 '13 at 10:40

You were almost there:

\begin{align} 1 &= \frac{\tan \alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\\ 1 - \tan\alpha\tan\beta &= \tan\alpha + \tan\beta\\ 2 &= 1 + \tan\alpha + \tan\beta + \tan\alpha\tan\beta\\ 2 &= (1+\tan\alpha)(1+\tan\beta) \end{align}

where each equation is equivalent to the preceding/following, and the two can be transformed into each other by a simple step.

• First multiply with the denominator (that's $\neq 0$),
• then add $1 + \tan\alpha\tan\beta$ to both sides,
• then write $1 + x + y + xy$ as the product $(1+x)(1+y)$.
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Thank you. I had got as far as $1 - \tan\alpha\tan\beta = \tan\alpha + \tan\beta$ but didn't think to rearrange to $1 = \tan\alpha +\tan\beta +\tan\alpha\tan\beta$ and then add 1 to both sides. – mikoyan Jul 18 '13 at 11:49
Always keep an eye on where you want to get. Here, you have $(1 + a)(1 + b) = 1 + a + b + ab$ as one side of the target. With practice, you recognise such patterns most of the time they occur, and then it becomes easy. – Daniel Fischer Jul 18 '13 at 11:55
I didn't downvote your solution, but I guess that the downvoter is displeased by the fact that you have just written four equations on top of each other. But you haven't indicated how they are connected with each other logically. Is one a consequence of another? Or are some two perhaps equivalent to each other? It is difficult to deduce anything with certainty, when such information is missing. – Jyrki Lahtonen Jul 19 '13 at 11:56
@JyrkiLahtonen Ah, that could be, thanks. I wish the downvoter had explained the reasoning, but well. Each equation follows from the preceding (or following, they're equivalent) by a simple transformation step. – Daniel Fischer Jul 19 '13 at 12:02
I realize that. But it is the common feature shared by your answer and that of iostream007's below. 10-15 years ago it really rubbed me the wrong way, when incoming freshmen did that (I was taught at an early age to use implication and equivalence arrows, and couldn't read what my students wrote at first). After a few years of struggle I have given up, but I still try to teach the kids to do the right thing :-) – Jyrki Lahtonen Jul 19 '13 at 12:20

You've reached here : $$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$$ Let's continue it in this way: \large\begin{align} \Rightarrow & \tan\alpha + \tan\beta+\tan\alpha\tan\beta = 1\\ \Rightarrow &\tan\alpha(1+\tan\beta) + \tan\beta = 1\\ \Rightarrow &\tan\alpha(1+\tan\beta) + 1+ \tan\beta = 2\\ \Rightarrow &(1 + \tan\alpha)(1 + \tan\beta) = 2\\ \end{align}

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@BabakS. Chashm. Manam kheili mohtaje doam. Lotfan mano az yad nabarin. mamnun – Mahdi Khosravi Jul 27 '13 at 18:16

You were almost there as pointed out by @Daniel in his answer.

Here's another way to do it:

$\beta=\dfrac{\pi}{4}-\alpha\implies (1+\tan\beta)=$ $\left(1+\tan\left(\dfrac{\pi}{4}-\alpha\right)\right)=\left(1+\dfrac{1-\tan\alpha}{1+\tan\alpha}\right)=\left(\dfrac{2}{1+\tan\alpha}\right)$

Thus, $$(1+\tan\beta)=\frac{2}{1+\tan\alpha}$$ $$\implies (1+\tan\alpha)(1+\tan\beta)=2$$

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from where OP left his step:$$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$$ $$\implies{1- \tan\alpha\tan\beta}=\tan\alpha + \tan\beta$$ $$\implies \tan\alpha + \tan\beta+\tan\alpha\tan\beta=1$$ add 1 to both sides $$\implies\tan\alpha + \tan\beta+\tan\alpha\tan\beta+1=2$$ $$\implies1+\tan\alpha + \tan\beta+\tan\alpha\tan\beta=2$$ factor the above equation $$\implies1(1+\tan\alpha) + \tan\beta(1+\tan\alpha)=2$$ $$\implies(1+\tan\alpha)(1+\tan\beta)=2$$ Hence proven

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somebody has mistaken my sentences so I am editing it – iostream007 Jul 18 '13 at 11:42
I didn't downvote your solution, but I guess that the downvoter is displeased by the fact that you have written seven sentences on top of each other, but you haven't indicated how they are connected with each other logically. Is one a consequence of another? Or are some two perhaps equivalent to each other? It is difficult to deduce anything with certainty, when such information is missing. I mean, the accepted answer has also received a downvote, and it suffers from the same malady. – Jyrki Lahtonen Jul 19 '13 at 11:53
@JyrkiLahtonen actually I just continue the asker's steps to proove it.Just doing some airthematic adjustmants and reach final step.I didn't found any thing tough to explain.can you mention any step which is not clear?Thanks – iostream007 Jul 19 '13 at 12:27
Your steps are all clear (that's why I didn't downvote). But I was raised to think that it is simply wrong not to add the equivalence arrows. I'm not the only one thinking that way either! But years of experience teaching mathematically talented youth fresh out of high school have convinced me that they are taught differently these days :-). See the comments under the accepted answer (and +1 to you). – Jyrki Lahtonen Jul 19 '13 at 12:45
@JyrkiLahtonen Yeah you are right about to add implication sign in each step In India we also used to learn this thing but here I didn't write cause I take new step to write new expression.And I just want to show the method.this is not my exam so I should worry about it.I am updating it in my answer thanks – iostream007 Jul 19 '13 at 22:43