Show that [(1+sinA-cosA)/(1+sinA+cosA)]^2 = (1-cosA)/(1+cosA)
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Step-by-step explanation:
Hi ,
LHS = [(1+sinA-cosA )/(1+sinA+cosA)]²
=[ (1+sin²A+cos²A+2sinA-2sinAcosA-2cosA)/(1+sin²A+cos²A+2sinA+2sinAcosA+2cosA) ]²
= [( 1+1+2sinA-2sinAcosA-2cosA)/(1+1+2sinA+2sinAcosA+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
I hope this helps you.
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