Question

How do you verify `(tan^2x + 1) / tan^2x = csc^2x`?

Answer 1

Use the Pythagorean identity `tan^2x + 1 = sec^2x` to start the simplification on the left side.

Explanation:

`sec^2x/tan^2x =`

Recall that `sec^2x = 1/cos^2x`, and that `tan^2x = sin^2x/cos^2x`

`(1/cos^2x)/(sin^2x/(cos^2x)) =`

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

`1/sin^2x =`

Remember that `1/sinx = cscx`

`csc^2x =`

Identity proved!!!

Hopefully this helps!

Answer 2

The definition of tangent is such that

`tan(x)=sin(x)/cos(x)`, then the left side of the equation is

`((sin(x)^2/cos(x)^2)+1)/(sin(x)^2/cos(x)^2)`

`=((sin(x)^2/cos(x)^2)+1)*cos(x)^2/sin(x)^2`

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

`=(sin(x)^2+cos(x)^2)/sin(x)^2`

and, using the fundamental property of `sin` and `cos` that `sin(x)^2+cos(x)^2=1` we have

`(sin(x)^2+cos(x)^2)/(sin(x)^2) = 1/(sin(x)^2)=csc(x)^2`.