Why is this trigonometric identity true?

There is a very direct way to solve this. First, use the given equation to find, for example, $\sin^2 \alpha$:

$$x=\sin^2 \alpha$$

$$\frac{1}{a} x^2+\frac{1}{b}(1-x)^2=\frac{1}{a+b}$$

Solving this easy quadratic, we get a single root:

$$x=\frac{a}{a+b}=\sin^2 \alpha$$

It follows:

$$1-x=\frac{b}{a+b}=\cos^2 \alpha$$

Now substitute these values into the second equation and see that it holds.

Solving the system $$\begin{cases}X^2+Y^2=1\\\dfrac{X^4}{a}+\dfrac{Y^4}{b}=\dfrac{1}{a+b}\end{cases}$$ where obviously $X$ and $Y$ are the sinus and cosinus respectively, one has $$X=-\sqrt{\dfrac{a}{a+b}}\\Y=\pm\sqrt{\dfrac{b}{a+b}}$$

This gives directly the equalities $\dfrac{X^8}{a^3}=\dfrac{a^4}{a^3(a+b)^4}$ and $\dfrac{Y^8}{b^3}=\dfrac{b^4}{b^3(a+b)^4}$ hence the result

$$\dfrac{X^8}{a^3}+\dfrac{Y^8}{b^3}=\dfrac{1}{(a+b)^3}$$

The general expression is:$$\frac{\sin^{4n}\theta}{a^{2n-1}} + \frac{\cos^{4n}\theta}{b^{2n-1}} = \frac{1}{(a+b)^{2n-1}}~, ~~n\in \mathbb N$$

From the given relation, it can be written \begin{align}(a+b)\left(\frac{\sin^4\theta}{a}+ \frac{\cos^4\theta}{b}\right) &= \left(\sin^2\theta + \cos^2\theta\right)^2\\ \implies\sin^4\theta + \cos^4\theta + \frac ba\sin^4\theta + \frac ab\cos^4\theta &= \sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta\\ \implies~~~~~~~~~~~ \left(\sqrt{\frac ba}\sin^2\theta - \sqrt\frac ab\cos^2\theta \right)^2&= 0\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\sin^2\theta}{a}&= \frac{\cos^2\theta }{b}\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\sin^2\theta}{a}= \frac{\cos^2\theta }{b} &= \frac{\sin^2\theta + \cos^2\theta}{a+b}\end{align}

From this, it can be concluded that $$\sin^2 \theta = \frac a{a+b};~~~\cos^2\theta = \frac{b}{a+b} \,.$$

It's a matter of substituting the values to prove the general and the desired expression.