Prove each idenity.
1+1/tan^2x=1/sin^2x
1/cosx-cosx=sinxtanx
1/sin^2x+1/cos^2x=1/sin^2xcos^2x
1/1-cos^2x+/1+cosx=2/sin^2x
and
(1-cos^2x)(1+1/tan^2x)= 1
I haven't even gotten 'round to sny of the quedtions because the first one is just so hard. I'm not really sure I'm uderstanding how to use the quotient and pythagorean identities. I'm so confused. I can't make sense of it and I have tried so many different ways. I must have spent over an hour on he first problem and still I can't come up with an answer. I'm just really frustrated; could someone please help me. I'd really appreciate it.
Sure, I'll walk you through the steps to prove each of the identities. Let's start with the first one:
1 + 1/tan^2(x) = 1/sin^2(x)
To prove this identity, we will work with the left-hand side (LHS) and simplify it until it equals the right-hand side (RHS).
Step 1: Start with the LHS and substitute tan^2(x) = sin^2(x)/cos^2(x), which is a Pythagorean identity.
LHS = 1 + 1/(sin^2(x)/cos^2(x))
Step 2: Simplify the expression by multiplying both sides of the fraction by cos^2(x).
LHS = 1 + cos^2(x)/sin^2(x)
Step 3: Find a common denominator for the two terms in the LHS.
LHS = (sin^2(x) + cos^2(x))/sin^2(x)
Step 4: Apply the Pythagorean identity sin^2(x) + cos^2(x) = 1.
LHS = 1/sin^2(x)
Now, we have obtained the RHS, which proves that 1 + 1/tan^2(x) = 1/sin^2(x).
Let's move on to the second identity:
1/cos(x) - cos(x) = sin(x)tan(x)
Step 1: Start with the LHS and simplify.
LHS = 1/cos(x) - cos(x)
Step 2: To perform subtraction with fractions, we need a common denominator. Multiply the first term by sin(x)/sin(x).
LHS = (1 * sin(x))/(cos(x) * sin(x)) - cos(x)
Step 3: Simplify the expression.
LHS = sin(x)/(cos(x)sin(x)) - cos(x)
Step 4: Apply the identity tan(x) = sin(x)/cos(x).
LHS = tan(x) - cos(x)
Now, we have obtained the RHS, which proves that 1/cos(x) - cos(x) = sin(x)tan(x).
I'll leave the remaining identities for you to try, but feel free to ask for help if you get stuck on any specific step.