(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

56 ur welcome: