三角比的關係

1/[1+(tanθ)^2] + (sin θ)^2
=1/[1+(sinθ/cosθ)^2] +(sin θ)^2
=1/{ [ (cos θ)^2+(sin θ)^2] / (cos θ)^2 } +(sin θ)^2
=1/[1/(cos θ)^2] +(sin θ)^2
=(cos θ)^2+ (sin θ)^2
=1

If you study in A. Math, there is a term secant which (sec θ)^2=1+(tan θ)^2 and cos θ=1/sec θ
1/[1+(tanθ)^2] + (sin θ)^2
=1/(secθ)^2 + (sin θ)^2
=(cos θ)^2+ (sin θ)^2
=1

Ans:
1/(1+tan^2 θ) + sin^2 θ
=1/[1+(sin^2θ/cos^2 θ)]+sin^2 θ
=1/[(cos^2 θ+sin^2 θ)/cos^2θ] + sin^2θ
=1/(1/cos^2 θ)+sin^2 θ
=cos^2 θ+sin^2 θ
=1//