Proof:

Let's start by simplifying the expression on the left side of the equation:

sinA - 2sin^3A = sinA - 2sinA(sin^2A)
= sinA(1 - 2sin^2A)
= sinA(cos^2A - sin^2A)
= sinA(cosA + sinA)(cosA - sinA)

Similarly, let's simplify the expression on the right side of the equation:

2cos^3A - cosA = cosA(2cos^2A - 1)
= cosA(2cosA + 1)(cosA - 1)

Now, let's rewrite the original equation using the simplified expressions:

(sinA - 2sin^3A) / (2cos^3A - cosA) = (sinA(cosA + sinA)(cosA - sinA)) / (cosA(2cosA + 1)(cosA - 1))

Canceling out common factors:

We can cancel out the common factors between the numerator and denominator:

(sinA(cosA + sinA)(cosA - sinA)) / (cosA(2cosA + 1)(cosA - 1))
= (sinA(sinA + cosA))/((2cosA + 1)(cosA - 1))

Using the identity sinA/cosA = tanA:

We can rewrite the expression as:

(sinA(sinA + cosA))/((2cosA + 1)(cosA - 1))
= tanA(sinA + cosA)/(2cosA + 1)(cosA - 1)

Using the identity sinA + cosA = √2sin(A + π/4):

We can rewrite the expression as:

tanA(sinA + cosA)/(2cosA + 1)(cosA - 1)
= tanA(√2sin(A + π/4))/(2cosA + 1)(cosA - 1)
= (tanA * √2sin(A + π/4))/(2cosA + 1)(cosA - 1)

Using the identity sinA/cosA = tanA:

We can rewrite the expression as:

(tanA * √2sin(A + π/4))/(2cosA + 1)(cosA - 1)
= (tanA * √2tan(A + π/4))/(2cosA + 1)(cosA - 1)
= (tanA * √2tan(A + π/4))/(2cosA + 1)(cosA - 1)

Using the identity tan(A + B) = (tanA + tanB)/(1 - tanA*tanB):

We can rewrite the expression as:

(tanA * √2tan(A + π/4))/(2cosA + 1)(cosA - 1)
= (tanA * √2(tanA + tan(π/4)))/(2cosA + 1)(cosA - 1)
= (tanA * √2(tanA + 1))/(2cosA + 1)(cosA - 1)

Using the identity tan(π