Prove that 1+sin2a/cos2a= sina+cos a/sina-cosa

1.	Prove that 1+sin2a/cos2a= sina+cos a/sina-cosa
Answer» $LHS = \frac{1+sin2A}{cos2A} \\= \frac{sin^{2}A+cos^{2}A+2sinAcosA}{cos^{2}A-sin^{2}A }$ $\underline { \blue { Using \: following \:formulae :}}$ Sin²A + Cos²A = 1 Sin2A = 2sinAcosA cos2A = cos²A - sin²A a²+b²+2ab = (a+b)² a²-b² = (a+b)(a-b) $= \frac{(sinA+cosA)^{2}}{(SinA+cosA)(sinA-cosA)} \\= \frac{sinA+cosA}{SinA-cosA}\\=RHS$ THEREFORE., $\red{\frac{1+sin2A}{cos2A}}\green {=\frac{sinA+cosA}{SinA-cosA}}$ •••♪

Answer»

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