<s>[INST] Is this a Viable/New Proof for Pythagoras Theorem? A triangle with side lengths $a$, $b$, $c$ and a height ($h$) that intercepts the hypotenuse ($c$) such that $c$ is split into two side lengths, $c = m + n$, we can find Pythagorean theorem using simple trigonometry.

We start by finding the sum of tan ratios that are equivalent,
$$\tan^{-1}\frac{a}{b}+\tan^{-1}\frac{b}{a} = 90^\circ$$
$$\tan^{-1}\frac{h}{m}+\tan^{-1}\frac{h}{n} = 90^\circ$$
$$\frac{a}{b}+\frac{b}{a}=\frac{h}{m}+\frac{h}{n}$$
Using $ab = hc$, Substitute $h = \dfrac{ab}{c}$,
$$\frac{a}{b}+\frac{b}{a}=\frac{ab}{cm}+\frac{ab}{cn}$$
After simplifying, we get:
$$\begin{align}
c&=\frac{ma^2b^2+na^2b^2}{mn\left(a^2+b^2\right)} \\[4pt]
&=\frac{(m + n)(a^2b^2)}{mn\left(a^2+b^2\right)} \\[4pt]
&=\frac{c(a^2b^2)}{mn\left(a^2+b^2\right)} \\[4pt]
mn &=\frac{a^2b^2}{\left(a^2+b^2\right)}
\end{align}$$
Note that $\sqrt{mn}=\dfrac{ab}{\sqrt{a^{2}+b^2}}$ and $h = \dfrac{ab}{c}$, this tells us that $\sqrt{mn} = h$.
$$\begin{align}
\frac{ab}{c}&=\frac{ab}{\sqrt{a^2+b^2}} \\[4pt]
c^2 &= a^2 + b^2
\end{align}$$
 [/INST] This amounts to a well-known similar-triangles argument, splitting the area, that $(a/c)^2+(b/c)^2=1$, which is a bit more direct.
As @DavidK notes, your opening use of trigonometry is invalid, and its conclusion (in fact, something stronger) should instead be deduced as per @WillOrrick's suggestion, giving $\tfrac{a}{b}=\tfrac{h}{m}$ and $\tfrac{b}{a}=\tfrac{h}{n}$, whence $1=\tfrac{h^2}{mn}=\tfrac{(ab)^2}{mnc^2}$, and the Pythagorean theorem is equivalent to $mn=\tfrac{(ab)^2}{a^2+b^2}$. You also use $\tfrac{a^2+b^2}{ab}=\tfrac{hc}{mn}=\tfrac{ab}{mn}$, which completes the proof.
So your strategy amounts to combining two formulae for area with$$u_1=u_2,\,v_1=v_2,\,u_1v_1=1\implies u_1+v_1=\tfrac{u_2+v_2}{u_2v_2}$$(where $u_1=v_1^{-1}=\tfrac{a}{b},\,v_1=\tfrac{h}{m},\,v_2=\tfrac{h}{n}$). But compare $\tfrac{a}{b}=\cot\theta$ with$$\left(\tfrac{a}{c}\right)^2=\tfrac{[\triangle CBH]}{[\triangle ABC]}=\tfrac{m}{m+n}=\tfrac{m/h}{v_1^{-1}+v_2^{-1}}=\tfrac{\cot\theta}{\cot\theta+\tan\theta}.$$Similarly (viz. $\theta\mapsto\tfrac{\pi}{2}-\theta$), $\left(\tfrac{b}{c}\right)^2=\tfrac{\tan\theta}{\cot\theta+\tan\theta}$. The proof technique I pointed to - that of seeing the Pythagorean theorem - notes the sum of these squares is $\tfrac{\color{blue}{blue}+\color{green}{green}}{\text{all}}=1$. Your proof works with fractions $k:=\cot\theta+\tan\theta$ times these, so Einstein's proof is equivalent to$$1=\tfrac{a/b+b/a}{k}=\tfrac{a^2+b^2}{kab}=\tfrac{a^2+b^2}{kch},$$where the last step uses your equation of two formulae for $[\triangle ABC]$, and the proof that $\tfrac{a^2+b^2}{kch}=1$ uses $kh=m+n=c$ by your $AB$-splitting observation that defines $m,\,n$.
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