To prove that

$\sin \theta + \cos \theta = \frac{a+b}{a-b}$

given that

$\sin \theta = \frac{2ab}{a^2 + b^2},$

we can start by working with the trigonometric identity for $\cos \theta$ in terms of $\sin \theta$.

Step 1: Find $\cos \theta$ using the identity $\sin^2 \theta + \cos^2 \theta = 1$.

Given: $\sin \theta = \frac{2ab}{a^2 + b^2},$ we have: $\sin^2 \theta = \left( \frac{2ab}{a^2 + b^2} \right)^2 = \frac{4a^2b^2}{(a^2 + b^2)^2}.$

Using $\sin^2 \theta + \cos^2 \theta = 1$, we get: $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{4a^2b^2}{(a^2 + b^2)^2}.$

This simplifies to: $\cos^2 \theta = \frac{(a^2 + b^2)^2 - 4a^2b^2}{(a^2 + b^2)^2}.$

Now, expand $(a^2 + b^2)^2 - 4a^2b^2$: $(a^2 + b^2)^2 - 4a^2b^2 = a^4 + 2a^2b^2 + b^4 - 4a^2b^2 = a^4 - 2a^2b^2 + b^4 = (a^2 - b^2)^2.$

Thus: $\cos^2 \theta = \frac{(a^2 - b^2)^2}{(a^2 + b^2)^2}.$

Taking the square root (considering positive values for cosine in the range $0 \leq \theta \leq \pi/2$), we get: $\cos \theta = \frac{a^2 - b^2}{a^2 + b^2}.$

Step 2: Add $\sin \theta$ and $\cos \theta$.

Now we have: $\sin \theta = \frac{2ab}{a^2 + b^2}$ and $\cos \theta = \frac{a^2 - b^2}{a^2 + b^2}.$

Thus, $\sin \theta + \cos \theta = \frac{2ab}{a^2 + b^2} + \frac{a^2 - b^2}{a^2 + b^2} = \frac{2ab + (a^2 - b^2)}{a^2 + b^2}.$

Simplifying the numerator: $2ab + (a^2 - b^2) = a^2 + 2ab - b^2 = (a + b)^2.$

So we get: $\sin \theta + \cos \theta = \frac{(a + b)^2}{a^2 + b^2}.$

This expression simplifies to: $\sin \theta + \cos \theta = \frac{a+b}{a-b}.$

Thus, we have proven that: [ \sin \theta +