Question

if m=sina+cosa and n=sina-cosa then find m^2-n^2

Answer 1

m = sinA + cosA


Square on both sides,

=> m² = ( sinA + cosA )²

$\mathbf{ => m ^{2} = sin ^{2} A + cos ^{2} A + 2sinAcosA} \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$



And,

n = sinA - cosA


Square on both sides,

=> n² = ( sinA - cosA )²

$\mathbf{=> n ^{2} = sin ^{2} A + cos ^{2} A - 2sinAcosA} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(ii)$



Then,

m² - n²

=> sin²A + cos²A + 2sinAcosA - sin²A - cos²A + 2sinAcosA

=> 4sinAcosA

Answer 2

Heya !!



m = Sin A + Cos A


n = Sin A - Cos A





m² - n²


=> ( SinA + Cos A )² - ( Cos A - Sin A )²




=> 4 Sin A Cos A [Since ( a +b )² - ( a - b)² = 4ab]