Question

Question #bca2e

Answer 1

It doesn't. A counterexample is `x=pi/6`

Explanation:

One form of the Pythagorean Identity states that `cot^2x+1=csc^2x`

If `cot^2x+sin^2x=csc^2x` then by the transitive property, `cot^2x+sin^2x=cot^2x+1`

Subtracting `cot^2x` from both sides,
`sin^2x=1`

If, say, `x=pi/6` then `sin^2x=1/4!=1`, and since we can provide a counterexample the identity is not true.

Plugging this counterexample into the original identity, we get `cot^2(pi/6)+sin^2(pi/6)=3.25` on the left side and `csc^2(pi/6)=4`. Obviously `3.25!=4` so the identity cannot be true.

Answer 2

`sin^2x=1` in the end, hence both sides of the equation are accounted for.
Read on for more info...

Explanation:

`cot^2x+sin^2x=csc^2x`

Since `csc^2x=1+cot^2x`,
`cot^2x+sin^2x=1+cot^2x`
`cot^2x-cot^2x+sin^2x=1`
`sin^2x=1`

Therefore,
`cot^2x+sin^2x=cot^2x+1=csc^2x`