Prove that sin(90-theta).cos(90-theta)/tan theta = 1-sin theta

First of all,we need to understand that:

theta(θ)

(sin90°-θ)=cosθ

(cos90°-θ)=sinθ

tanθ=sinθ/cosθ

1/tanθ=cosθ/sinθ

cosθ^2+sinθ^2=1

In this case:

sin(90°-θ) × cos(90°-θ)/tanθ
=cosθ × sinθ × cosθ/sinθ
=cosθ × cosθ
=cos^2 θ
=1-sin^2 θ

Therefore, we can see that:
sin(90°-θ) × cos(90°-θ)/tanθ = 1-sin^2θ

Hope it helps!