Question

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

Answer 1

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

$\underline { \blue { Using \: following \:formulae :}}$

• 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.,

$\red{\frac{1+sin2A}{cos2A}}\green {=\frac{sinA+cosA}{SinA-cosA}}$

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