Question

Question #49155

Answer 1

By Pythagorean identity we have

`sin^2x+cos^2x=1`

Dividing both sides by `sin^2x` we get

`sin^2x/sin^2x+cos^2x/sin^2x=1/sin^2x`

`=>1+cot^2x=csc^2x`

`=>csc^2x-cot^2x=1`