Question

(1+sinA+cosA)^2 = 2(1+sinA) (1+cosA)

Answer 1

Answer:

Step-by-step explanation:

$L.H.S = 1^2+Sin^2A+Cos^2A+2SinA+2CosA+2SinACosA\\ \{(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\}\\ = 1 + (Sin^2A+Cos^2A) + 2SinA+2CosA+2SinACosA \\ \{Sin^2A+Cos^2A=1\}\\ = 1 + 1 + 2SinA+2CosA+2SinACosA \\ = 2(1 + SinA+CosA+SinACosA)\\ = 2(1^2 + (SinA+CosA).1+ SinACosA )\\ \{x^2+(a+b)x + ab = (x+a)(x+b)\} \quad here \quad x=1\\ = 2(1+SinA)(1+CosA).$