plz...三角學練習θ

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

b.(1-cosθ)(1+cosθ)/tan^2θ
=(1-cos^2θ)/tan^2θ
=sin^2θ/tan^2θ
=cos^2θ

c.己知cos θ=3/5,其中θ是一個銳角,求sinθ及tanθ的值
sinθ
=√(1-cos^2θ)
=√(1-(3/5)^2)
=4/5
tanθ
=sinθ/cosθ
=(4/5) / (3/5)
=4/3