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My niece came to me with a math question for her schoolbooks: Proof the equation below, when you know a + b = 45°: (1+tan(a))(1+tan(b)) = 2

It seemed a pretty easy equation to me so I thought I'd give it a go. Ended up changing the 1 with tan(a+b) and then using tan(a+b)=(tan(a)+tan(b))/(1-tan(a).tan(b)) , but that kinda got me stuck...

Anyone knows what the correct approach is here?

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    $\begingroup$ You need to remember the tangent of $45$ degrees. $\endgroup$ Oct 14, 2018 at 7:56
  • $\begingroup$ Yes, that's why I changed the 1 with tan(a+b) . Or should I do anything else with it? $\endgroup$
    – binoculars
    Oct 14, 2018 at 8:20

2 Answers 2

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It's not an equation, since there's no unknown. It's only a formula to prove.

Hint:

Expand $(1+\tan a)(1+\tan b)$ and use the addition formula for the tangent: $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\,\tan b}.$$

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  • $\begingroup$ Got it! Thanks a lot!!! $\endgroup$
    – binoculars
    Oct 14, 2018 at 8:26
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You have\begin{align}(1+\tan a)(1+\tan b)&=(1+\tan a)\bigl(1+\tan(45^\circ-a)\bigr)\\&=(1+\tan a)\left(1+\frac{1-\tan a}{1+\tan a}\right)\\&=1+\tan(a)+1-\tan(a)\\&=2.\end{align}

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  • $\begingroup$ Thanks! Could you maybe explain how you go from tan(45°-a) to (1-tan(a))/(1+tan(a))? $\endgroup$
    – binoculars
    Oct 14, 2018 at 8:23
  • $\begingroup$ The formula for $\tan(a-b)$ is simply the addition formula, with the $+$ and $-$ signs in the numerator and denominator swapped, and $\tan 45^\circ=1$. $\endgroup$
    – Bernard
    Oct 14, 2018 at 8:31
  • $\begingroup$ \begin{align}\tan(x-y)&=\tan\bigl(x+(-y)\bigr)\\&=\frac{\tan(x)+\tan(-y)}{1-\tan(x)\tan(-y)}\\&=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}.\end{align} $\endgroup$ Oct 14, 2018 at 8:42

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