Given that $\alpha$ and $\beta$ are acute angles, show that $\alpha+\beta = \pi/2$ if and only if $\cos^2{\alpha} +\cos^2{\beta} = 1$. This question is from an exam paper:

Given that $\alpha$ and $\beta$ are acute angles, show that $\alpha+\beta = \pi/2$ if and only if $\cos^2{\alpha} +\cos^2{\beta}  = 1$.

I want to do it in one line rather than doing both directions separately. Is this a valid proof?
$\cos^2{\alpha}+\cos^2{\beta} =\cos^2{\alpha} +\sin^2(\pi/2-\beta)=\cos^2{\alpha}+\sin^2{\alpha}=1$
Edit: Thanks, I now realise I've only proven the 'only if' part. Here is my attempt at the 'if' part, which was more complicated than I thought it would be.
$\begin{align}
&&\cos^2{\alpha} +\cos^2{\beta}  = \cos^2{\alpha} +\sin^2(\pi/2-\beta)=1\\
\implies&&\sin^2(\pi/2-\beta)=\sin^2{\alpha}\\
\implies&&\pi/2-\beta=\pm\alpha+\pi n\\
\implies&&\alpha+\beta=\pi/2+\pi n\end{align}$
$\alpha$ and $\beta$ are acute so $0<\alpha + \beta<\pi$ so $n=0$.
 A: The condition that $\alpha$ and $\beta$ are acute implies that the cosines are positive, then $\cos^2{\alpha} +\cos^2{\beta}  = 1$ implies $\cos{\alpha} +\cos{\beta} \ge 1$. 
Hence we can construct a triangle with sides $1,\cos{\alpha},\cos{\beta}$. 
Now, by Pythagoras' theorem, either of the statements $\alpha+\beta = \pi/2$ and $\cos^2{\alpha} +\cos^2{\beta}  = 1$ is equivalent to the fact that this triangle is right-angled.
A: It's impossible to do both directions in one line since at some point you're going to have to use words/other facts to explain why them being acute angles forces $\alpha+\beta=\pi/2$ (as opposed to some other relation that would also make the sum of the squares equal $1$).
Hint: Except for the wrinkle above, the other direction can look similar, though.
A: HINT:
For the converse, we have $$0=\cos^2\alpha-(1-\cos^2\beta)=\cos^2\alpha-\sin^2\beta=\cos(\alpha+\beta)\cos(\alpha-\beta)$$
As $\displaystyle 0<\alpha,\beta<\frac\pi2, \alpha\sim\beta<\frac\pi2\implies  \cos(\alpha-\beta)>0 $
A: The other direction
$$\begin{cases}\cos^2\alpha+\cos^2\beta=1=\cos^2\alpha+\sin^2\alpha\implies\cos^2\beta=\sin^2\alpha\\{}\\\cos^2\alpha+\cos^2\beta=1=\sin^2\beta+\cos^2\beta\implies\cos^2\alpha=\sin^2\beta\end{cases}\;\implies$$
$$\cos\beta=\sin\alpha>0\;,\;\;\sin\beta=\cos\alpha>0$$
since both cosine and sine positive in the first quadrant , so
$$\sin\beta\cos\beta=\sin\alpha\cos\alpha\iff \sin2\beta=\sin 2\alpha$$
so that
$$2\alpha=2\beta\iff \alpha=\beta\;,\;\;\text{or}\;\;2\alpha=\pi-2\beta$$
and we're done (Because $\;\alpha=\beta\;$ and $\;\cos^2\alpha+\cos^2\beta=1\implies \alpha=\beta=\frac\pi4\;$ and everything's fine)
A: Consider the following arrangements within the unit circle $O$, where $OC$ and $XY$ are (congruent) diagonals of rectangle $\square OXCY$:

$$\begin{align}
\cos^2\alpha + \cos^2\beta = 1 \quad &\Leftrightarrow \quad |XY| = 1 \\
&\Leftrightarrow \quad |OC| = 1 = |OA| = |OB| \\
&\Leftrightarrow \quad A = B = C \\
&\Leftrightarrow \quad \alpha + \beta = \pi/2
\end{align}$$
