Prove that \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}

1.	Prove that \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta} )^{2}=\frac{1-\cos \theta}{1+\cos \theta} 1+sin 6 - 0056 ) " 1-cos®1+sin+cos® ) 1+cos®
Answer» 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 ) ] = RHS

Prove that \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta} )^{2}=\frac{1-\cos \theta}{1+\cos \theta} 1+sin 6 - 0056 ) " 1-cos®1+sin+cos® ) 1+cos®

Answer»

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 ) ]

= RHS