Proof that (cos3x-sin3x)/(1−2sin2x) =cosx+sinx

$\cos 3x = 4\cos^3 x - 3\cos x$
$\sin 3x = 3\sin x - 4\sin^3 x$

Substitute these into the numerator of the LHS:

$$\text{Numerator} = (4\cos^3 x - 3\cos x) - (3\sin x - 4\sin^3 x)$$ $$\text{Numerator} = 4\cos^3 x - 3\cos x - 3\sin x + 4\sin^3 x$$ $$\text{Numerator} = 4(\cos^3 x + \sin^3 x) - 3(\cos x + \sin x)$$

$$\cos^3 x + \sin^3 x = (\cos x + \sin x)(\cos^2 x - \cos x \sin x + \sin^2 x)$$

Since $\cos^2 x + \sin^2 x = 1$, this simplifies to:

$$\cos^3 x + \sin^3 x = (\cos x + \sin x)(1 - \cos x \sin x)$$

$$\text{Numerator} = 4(\cos x + \sin x)(1 - \cos x \sin x) - 3(\cos x + \sin x)$$

Factor out $(\cos x + \sin x)$:

$$\text{Numerator} = (\cos x + \sin x) [4(1 - \cos x \sin x) - 3]$$ $$\text{Numerator} = (\cos x + \sin x) [4 - 4\cos x \sin x - 3]$$ $$\text{Numerator} = (\cos x + \sin x) [1 - 4\cos x \sin x]$$

$$\text{Denominator} = 1 - 2\sin 2x$$ $$\text{Denominator} = 1 - 2(2\sin x \cos x)$$ $$\text{Denominator} = 1 - 4\sin x \cos x$$

$$\frac{\cos 3x - \sin 3x}{1 - 2\sin 2x} = \frac{(\cos x + \sin x) [1 - 4\cos x \sin x]}{1 - 4\sin x \cos x}$$

Assuming $1 - 4\cos x \sin x \neq 0$, we can cancel the common term in the numerator and denominator.

$$\frac{\cos 3x - \sin 3x}{1 - 2\sin 2x} = \cos x + \sin x$$

This matches the RHS.

Thus, the identity is proven.