Question

1 - cos2A/1 + sin2A = sin A

Answer 1

Answer:

tanA

Step-by-step explanation:

Given:\frac{1-cos2A+sin2A}{1+cos2A+sin2A}Given:

1+cos2A+sin2A

1−cos2A+sin2A

∴ cos 2A = (1 - sin²A)

∴ cos 2A = 2 cos²A - 1

∴ sin 2A = 2 sin A cos A

=\frac{1-(1-2sin^2A)+2sinAcosA}{1+(2cos^2A-1)+2sinAcosA}=

1+(2cos

2

A−1)+2sinAcosA

1−(1−2sin

2

A)+2sinAcosA

=\frac{1-1+2sin^2A+2sinAcosA}{1+2cos^2A-1+2sinAcosA}=

1+2cos

2

A−1+2sinAcosA

1−1+2sin

2

A+2sinAcosA

=\frac{2sin^2A+2sinAcosA}{2cos^2A+2sinAcosA}=

2cos

2

A+2sinAcosA

2sin

2

A+2sinAcosA

=\frac{2sinA(sinA+cosA)}{2cosA(sinA+cosA)}=

2cosA(sinA+cosA)

2sinA(sinA+cosA)

=\frac{sinA}{cosA}=

cosA

sinA

=\boxed{tanA}=

tanA