Verified answer

(1 + sin 2θ - cos 2θ) / (1 +

sin 2θ + cos 2θ) = tan θ

sin 2θ = 2sinθcosθ

cos 2θ = 2cos^2 θ - 1

cos^2 θ + sin^2 θ = 1

substituting, we have,

= (1+2sinθcosθ-2cos^2 θ+1) / (1+2sinθcosθ+2cos^2 θ -1)

= (2+2sinθcosθ-2cos^2 θ) / (2sinθcosθ-2cos^2 θ)

= (2(cos^2 θ + sin^2 θ)+2sinθcosθ-2cos^2 θ) / [2cosθ(sinθ-cosθ)]

=(2cos^2 θ + 2sin^2 θ + 2sinθcosθ - 2cos^2 θ) / (2cosθ(sinθ-cosθ))

=(2sin^2 θ + 2sinθcosθ)/[2cosθ(sinθ-cosθ)]

=[2sinθ(sinθ-cosθ)]/[2cosθ(sinθ-cosθ)]

sinθ-cosθ cancels out:

=2sinθ/2cosθ

=sinθ/cosθ=tanθ. Proven.

LHS

= (1 + sin 2θ - cos 2θ) / (1 + sin 2θ + cos 2θ)

= [sin2θ + (1 - cos2θ)] / [sin2θ + (1 + cos2θ)]

= [2sinθ cosθ + 2sin^2 θ] / [2sinθ cosθ + 2cos^2 θ]

= [sinθ (cosθ + sinθ)] / [cosθ (sinθ + cosθ]

= sinθ/cosθ

= tanθ

= RHS

(1 + sin(2t) - cos(2t)) / (1 + sin(2t) + cos(2t)) =>

(1 + sin(2t) + cos(2t) - 2 * cos(2t)) / (1 + sin(2t) + cos(2t)) =>

(1 + sin(2t) + cos(2t)) / (1 + sin(2t) + cos(2t)) - 2 * cos(2t) / (1 + sin(2t)+ cos(2t)) =>

1 - 2 * cos(2t) / (1 + sin(2t) + cos(2t)) =>

1 - 2 * (cos(t)^2 - sin(t)^2) / (1 + 2sin(t)cos(t) + cos(t)^2 - sin(t)^2) =>

1 - 2 * (cos(t) - sin(t)) * (cos(t) + sin(t)) / (cos(t)^2 + cos(t)^2 + 2sin(t)cos(t)) =>

1 - 2 * (cos(t) - sin(t)) * (cos(t) + sin(t)) / (2cos(t)^2 + 2sin(t)cos(t)) =>

1 - 2 * (cos(t) - sin(t)) * (cos(t) + sin(t)) / (2cos(t) * (cos(t) + sin(t)) =>

1 - (cos(t) - sin(t)) / cos(t) =>

1 - (cos(t)/cos(t)) + sin(t)/cos(t) =>

1 - 1 + tan(t) =>

tan(t)