.
sec(270∘−θ)sec(90∘−θ)−tan(270∘−θ)tan(90∘+θ)=−1
1cos(270∘−θ)⋅1cos(90∘−θ)−sin(270∘−θ)cos(270∘−θ)⋅sin(90∘+θ)cos(90∘+θ)=−1
We have the following four identities:
sin(α+β)=sinαcosβ+cosαsinβ
sin(α−β)=sinαcosβ−cosαsinβ
cos(α+β)=cosαcosβ−sinαsinβ
cos(α−β)=cosαcosβ+sinαsinβ
Therefore,
cos(270∘−θ)=cos270∘cosθ+sin270∘sinθ=(0)cosθ+(−1)sinθ=0−sinθ=−sinθ
cos(90∘−θ)=cos90∘cosθ+sin90∘sinθ=(0)cosθ+(1)sinθ=0+sinθ=sinθ
sin(270∘−θ)=sin270∘cosθ−cos270∘sinθ=(−1)cosθ−(0)sinθ=−cosθ−0=−cosθ
sin(90∘+θ)=sin90∘cosθ+cos90∘sinθ=(1)cosθ+(0)sinθ=cosθ+0=cosθ
cos(90∘+θ)=cos90∘cosθ−sin90∘sinθ=(0)cosθ−(1)sinθ=0−sinθ=−sinθ
Now, let's substitute all the pieces:
1−sinθ⋅1sinθ−−cosθ−sinθ⋅cosθ−sinθ=−1sin2θ+cos2θsin2θ=
−1+cos2θsin2θ=−(1−cos2θ)sin2θ=−sin2θsin2θ=−1
