If $\alpha+\beta=\pi/4$, then simplify $(\tan\alpha+1)(\tan\beta+1)$.

Question

This was a question asked in an exam to qualify for a Monbukagakusho scholarship in 2016. I decided to give it a go yet I still don't really know how to solve the equation. I tried solving for alpha and beta but I don't think that is needed to answer the problem. If someone could give me a hint on how to solve this problem I would appreciate.

If $\alpha$, $\beta$ are numbers satisfying $0<\alpha<\frac{\pi}{4}$, and $0<\beta<\frac{\pi}{4}$, and $\alpha+\beta=\frac{\pi}{4}$, then it follows that $$(\tan\alpha+1)(\tan\beta+1) = \text{???}$$

Answer 1

Using the hint from the comments, we have

$$1=\tan(\alpha+\beta)={\tan\alpha+\tan\beta\over1-\tan\alpha\tan\beta}$$

so $\tan\alpha+\tan\beta=1-\tan\alpha\tan\beta$. It follows that

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

Remark: A quick way to arrive at $2$ as the answer is to note that if there is an answer, then it should hold for all $\alpha$ and $\beta$ subject to $\alpha+\beta=\pi/4$, hence in particular for $\alpha=0$ and $\beta=\pi/4$, for which $(\tan\alpha+1)(\tan\beta+1)=(0+1)(1+1)=2$. (If you want to adhere to the limits $0\lt\alpha,\beta\lt\pi/4$, then take the limit as $(\alpha,\beta)\to(0,\pi/4)$.)

Answer 2

First expand, then get rid of $\beta$: $$ \begin{align} (\tan\alpha+1)(\tan\beta+1)&=\tan\alpha\tan\beta+\tan\alpha+\tan\beta+1\\ &=\tan\alpha\tan\left(\frac{\pi}{4}-\alpha\right)+\tan\alpha+\tan\left(\frac{\pi}{4}-\alpha\right)+1 \end{align} $$ Now solve for $\tan\left(\frac{\pi}{4}-\alpha\right)$: $$ \begin{align} \tan\left(\frac{\pi}{4}-\alpha\right)&=\frac{\tan\frac{\pi}{4}-\tan\alpha}{1+\tan \frac{\pi}{4}\tan\alpha}\\ &=\frac{1-\tan\alpha}{1+\tan\alpha} \end{align} $$ Plug this in to the original equation and we're done.

Answer 3

The hint given by Mark should have helped you a lot. By the way I'm giving you the answer:

if $a + b= \pi/4$ then $a = \pi/4 - b$.

So

$(\tan(a) +1)(\tan(b) +1)\\= (\tan(\pi/4-b) +1)(\tan(b) +1)\\=\dfrac{(1-\tan(b)+1-\tan(b))}{(1+\tan(b))}(\tan(b)+1)\\=2.$

Answer 4

The simplest solution is to observe $$1 = \tan \frac{\pi}{4} = \tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan\alpha \tan\beta},$$ therefore $$1 - \tan \alpha \tan \beta = \tan \alpha + \tan \beta,$$ or $$1 = \tan \alpha \tan \beta + \tan \alpha + \tan \beta,$$ or $$2 = \tan \alpha \tan \beta + \tan \alpha + \tan \beta + 1 = (\tan \alpha + 1)(\tan \beta + 1).$$