How to prove identity for the expression

Question

What are the steps and transformations for the expression:

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

Answer 1

in "short hand" call cos(x)=x, and sin(x)=y,,,,,,x²+y²=1

want to show

(1+x²)/y²=(2/y²)-1

1+x²=2-y²

x²+y²=1

1=1

in actual trig

1+cos²(x)=2-sin²(x),,,,multiplying both sides by sin²(x)

cos²(x)+sin²(x)=1

1=1

Answer 2

Begin with (1 + cos²(x))/sin²(x) and split it apart into two fractions:

1/sin²(x) + cos²(x)/sin²(x)

Using 1/sin(x) = csc(x) and cos(x)/sin(x) = cot(x), rewrite this as:

csc²(x) + cot²(x)

Using the Pythagorean Identity cot²(x) + 1 = csc²(x) subtract 1 from both sides to get cot²(x) = csc²(x) - 1 and then replace cot²(x) with csc²(x) - 1:

csc²(x) + csc²(x) - 1

Combine like terms to get:

2csc²(x) - 1