請幫忙回答中二數學題

1. tan θ/(1+ tan^2 θ)
= [(sinθ)/(cosθ)] / (1+ tan^2 θ) ~~ {因為 tanθ = sinθ/cosθ}
= [(sinθ)/(cosθ)] / (cos^2 θ/cos^2 θ + sin^2 θ/cos^2 θ) ~~ {在分母部分用cos^2 θ通分母}
= [(sinθ)/(cosθ)] / (1/cos^2 θ) ~~ {因為 cos^2 θ + sin^2 θ = 1}
= [(sinθ)/(cosθ)]x[(cos^2 θ)]
= (sinθ)(cosθ)

2. cos(40° - 3θ) = sin(7θ + 2° )
sin [90°- (40° - 3θ)] = sin(7θ + 2° )
sin [50°- 3θ] = sin(7θ + 2° )
所以 [50°- 3θ] = (7θ + 2° )
4θ = 48°
θ = 12°

3. (sin 19° tan 71° - cos 19° )^2
= [sin 19° 1/tan (90°- 71°) - cos 19° ]^2
={[sin 19° (cos19°)/(sin 19°) - cos 19° ]^2
={[ (cos19°) - cos 19° ]^2
= 0^2
= 0

1. LHS:
tan θ/(1+ tan^2 θ)
= tan θ/(1+ sin^2 θ/cos^2 θ)
= tan θ/((cos^2 θ + sin^2 θ)/cos^2 θ)
= tan θ/(1/cos^2 θ)
= (sinθ/cosθ)*cos^2θ
= sinθcosθ
= RHS

3. 5.4321 (use calculator)