(partially done! ) verify the identity and show your work.
cot^2x + csc^2x = 2csc^2x – 1.
my teacher told me to use cot^2(x)+1 = csc^2(x) as a starting point and must show that lhs=rhs.

Throughout all of these steps I'm only going to alter the left hand side (LHS). I am NOT going to change the right hand side (RHS) at all.

Before I change the LHS of the original equation, let's focus on the given identity
cot^2(x) + 1 = csc^2(x)

Since we know it's an identity, we can subtract 1 from both sides and the identity would still hold true
cot^2(x) + 1 = csc^2(x)
cot^2(x) + 1-1 = csc^2(x)-1
cot^2(x) + 0 = csc^2(x)-1
cot^2(x) = csc^2(x)-1

So we'll use the identity cot^2(x) = csc^2(x)-1



Now onto the main equation given

cot^2(x) + csc^2(x) = 2csc^2(x) - 1
cot^2(x) + csc^2(x) = 2csc^2(x) - 1 note the term in bold
csc^2(x)-1 + csc^2(x) = 2csc^2(x) - 1 note the terms in bold
[ csc^2(x) + csc^2(x) ] - 1 = 2csc^2(x) - 1
[ 2csc^2(x) ] - 1 = 2csc^2(x) - 1
2csc^2(x) - 1 = 2csc^2(x) - 1

The bold terms indicate how the replacements occur.

So the original equation has been proven to be an identity because the LHS has been altered to transform into the RHS