問2題數學題,快d黎啦!!!!!!!!!!!!!!!!!!!!!!!!

Ans:
1. 1/sinθ- sinθ/tan^2θ
=1/sinθ- sinθ/(sin^2θ/cos^2θ)
=1/sinθ- sinθ(cos^2θ/sin^2θ)
=1/sinθ- (cos^2θ/sinθ)
=(1-cos^2θ)/sinθ
=sin^2θ/sinθ
=sinθ

2. (sinθ cosθ)^2-1
=(sinθ)^2 2sinθcosθ (cosθ)^2 -1
= 2sinθcosθ 1-1
= 2sinθcosθ

答案:
1. 1/sinθ- sinθ/(tanθ)^2

=1/sinθ- sinθ/((sinθ)^2/(cosθ)^2)

=1/sinθ- sinθ((cosθ)^2/(sinθ)^2)

=1/sinθ- ((cosθ)^2/sinθ)

=(1-(cosθ)^2)/sinθ

=((sinθ)^2 + (cosθ)^2 -cos^2θ)/sinθ(因為(sinθ)^2 + (cosθ)^2=1)

=(sinθ)^2/sinθ

=sinθ

2. (sinθ+cosθ)^2-1

=(sinθ)^2+2sinθcosθ+(cosθ)^2 -1

= 2sinθcosθ+1-1(因為(sinθ)^2 + (cosθ)^2 = 1 , 要背嫁)

= 2sinθcosθ