Question

((1+sinA-cosA)/(1+sinA+cosA))²= 1-cosA/1+cosA​

Answer 1

Step-by-step explanation:

LHS=[(1+sinA - cosA) / (1+sinA + cosA)]²

=[(1+ sin²A+cos²A + 2sinA - 2sinAcosA-

2cosA) / (1+sin²A + cos²A + 2sin A+ 2sinA cosA + 2cosA)]²

=[(1+1+2sinA - 2sinA cosA - 2cosA) / (1+1+2sinA + 2sinA cosA + 2cosA)]²

={(2(1+sinA)-2cosA(sinA+1)/2(1+sinA)+2cosA (sinA+1)}]²

={(2(1+sinA)[1-cosA]/[2(1+sinA)(1+cosA)}]²

=[(1-cosA)/(1+cosA)]²

=RHS

LHS=RHS , PROVED

Answer 2

Step-by-step explanation:

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