Question

Prove that $\frac{cos 3A + sin 3A}{cos A - sin A}$ = 1 + 2 sin 2A.

Answer 1

$LHS=\frac{cos3A+sin3A}{cosA-sinA} \\\\=\frac{cos3A}{cosA-sinA}+\frac{sin3A}{cosA-sinA}\\\\=\frac{cos(2A+A)}{cosA-sinA}+\frac{sin(2A+A)}{cosA-sinA}$

$=\frac{cos2A.cosA-sin2A.sinA}{cosA-sinA}+\frac{sin2A.cosA+cos2A.sinA}{cosA-sinA}\\\\=\frac{cos2A.cosA-sin2A.sinA+sin2A.cosA+cos2A.sinA}{cosA-sinA}\\\\=\frac{cos2A.cosA+cos2A.sinA+sin2A.cosA-sin2A.sinA}{cosA-sinA}\\\\=\frac{cos2A.cosA+cos2A.sinA}{cosA-sinA}+\frac{sin2A.cosA-sin2A.sinA}{cosA-sinA}\\\\=\frac{cos2A(cosA+sinA)}{cosA-sinA}+\frac{sin2A(cosA-sinA)}{cosA-sinA}\\\\=\frac{cos2A(cosA+sinA)(cosA+sinA)}{(cosA-sinA)(cosA+sinA)}+sin2A$

$=\frac{cos2A(cos^{2}A+sin^{2}A+2sinAcosA)}{cos^{2}A-sin^{2}A}+sin2A\\\\=\frac{cos2A(1+sin2A)}{cos2A}+sin2A\\\\=1+sin2A+sin2A\\\\=1+2sin2A\\\\=RHS$