Question

How do you prove `cot^2 x +csc^2 x = 2csc^2 x - 1`?

Answer 1

See below.

Explanation:

I would rewrite in terms of sine and cosine.

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

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

Recall that `cos^2x + sin^2x = 1 -> sin^2x =1 - cos^2x`.

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

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

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

`LHS = RHS`

Proved!!

Hopefully this helps!

Answer 2

We know, `csc^2x -cot^2 x=1`

So,multiplying both side with `(-1)` we get,

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

Now, adding `2 csc^2 x` on both side of the equation we get,

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

So,`2 csc^2 x -1 = cot^2 x +csc^2x`

Proved