If $\frac{\sin^2 x}{a} = \frac{\cos^2 x}{b}$, then prove that they are also equal to $\frac{\sin^2 x + \cos^2 x}{a+b}$

Question

If $ \dfrac{\sin^2 x}{a} = \dfrac{\cos^2 x}{b}$, then prove that they are also equal to $\dfrac{\sin^2 x + \cos^2 x}{a+b}$.

I don't know how to solve this.

I was thinking about componeto dividento, but it doesn't help either.

Answer 1

It is generally true that $\frac a b =\frac c d$ implies $\frac a b =\frac c d=\frac {a+c} {b+d}$. For a proof let $r=\frac a b =\frac c d$, write $a=rb,c=rd$ and calculate $\frac {a+c} {b+d}$.

Answer 2

We will prove a proposition in general.

Proposition 1. If $a/b=c/d$ then they are also equal to $\frac{a\pm c}{b\pm d}$.

Proof. We know that $ad=bc$. So $$bc=ad$$ $$ab\pm bc= ab\pm ad$$ $$(a\pm c)b=a(b\pm d)$$ $$(a\pm c)/(b\pm d)=a/b$$ Actually I wrote these equations from bottom to top, and reversed them (since it is easier that way as you can see).

Answer 3

$${\dfrac{\sin^2 x}{a} = \dfrac{\cos^2 x}{b}\implies\\\dfrac{a}{\sin^2 x} = \dfrac{b}{\cos^2 x}\implies\\\dfrac{a}{\sin^2 x}+\dfrac{b}{\sin^2 x} = \dfrac{b}{\sin^2 x}+\dfrac{b}{\cos^2 x}\implies\\{a+b\over \sin^2 x}={b(\sin^2 x+\cos^2 x)\over \sin^2 x\cos^2 x}={b\over \sin^2 x\cos^2 x}\implies\\{a+b}={b\over \cos^2 x}\implies \\{\sin^2 x\over a}={\cos^2x\over b}={1\over a+b}={\sin^2x+\cos^2x\over a+b}}$$