Question

Show that

[1+1/tan^2A][1+1/cot^2A] = 1/sin^2A-sin^4A

Answer 1

[1+1/tan^2A][1+1/cot^2A]

= [(1 + tan^2A)/(tan^2A)][(1 + cot^2A)/(cot^2A)]
= (sec^2A/tan^2A)*(cosec^2A/cot^2A)
= 1/(cos^2A*sin^2A)
= 1/[(1 - sin^2A)(sin^2A)]
= 1/sin^2A-sin^4A