必修四同角三角函数关系求证：1/sin^2x+1/cos^2x-1/tan^2x=2+tan^2x

Question

必修四同角三角函数关系
求证：1/sin^2x + 1/cos^2x - 1/tan^2x =2 + tan^2x

Answer 1

(1/sin^2 x) + (1/cos^2 x) - （1/tan^2 x）= (1/sin^2 x*cos^2 x)-(cos^2 x/sin^2 x)
=(1-cos^4x)/sin^2 x*cos^2 x
=(1+cos^2x)sin^2x/sin^2 x*cos^2 x
==(1+cos^2x)/cos^2 x
=sec^2x+1=1+tan^2x+1= 2 + (tan^2 x)