For each expression in column I, choose the expression from column II to complete an identity:

Column I Column II

1. -tanxcosx A. sin^2x/cos^2x
2. sec^2x-1 B. 1/sec^2x
3. sec x/cscx C. sin(-x)
4. 1+sin^2x D.csc^2x-cot^2x+sin^2x
5. cos^2 x E. tanx

I figured #1 is C, #2 is A, And #4 is D, but I am not sure about the rest. Please give answers and explain. Also, for the following, verify that each trig equation is an identity and explain why. NOTE: I using @ to mean angle.

6. cot^2t-1/1+cot^2t = 1-2sin^2t
7. 1-cos@/1+cos@ = 2csc^2@-2csc@cot@-1

Let's go through each expression and match them with the corresponding identity:

1. -tan(x)cos(x) = C. sin(-x)
The negative of the tangent of x multiplied by the cosine of x is equal to the sine of the negative of x.

2. sec^2(x) - 1 = B. 1/sec^2(x)
The secant squared of x minus 1 is equal to the reciprocal of the secant squared of x.

3. sec(x)/csc(x) = E. tan(x)
The secant of x divided by the cosecant of x is equal to the tangent of x.

4. 1 + sin^2(x) = A. sin^2(x)/cos^2(x)
The quantity 1 plus the sine squared of x is equal to the sine squared of x divided by the cosine squared of x.

5. cos^2(x) = E. tan(x)
The cosine squared of x is equal to the tangent of x.

Now, let's verify the given trigonometric identities and explain why they are identities:

6. cot^2(t) - 1 / (1 + cot^2(t)) = 1 - 2sin^2(t)
To prove this identity, we start with the left-hand side (LHS):
LHS = cot^2(t) - 1 / (1 + cot^2(t))
Using the Pythagorean identity, cot^2(t) = 1 + csc^2(t), we can substitute and simplify:
LHS = (1 + csc^2(t)) - 1 / (1 + (1 + csc^2(t)))
= csc^2(t) / (2 + csc^2(t))
Applying the Pythagorean identity for sine, csc^2(t) = 1 + cot^2(t), we have:
LHS = (1 + cot^2(t)) / (2 + 1 + cot^2(t))
= 1 + cot^2(t) / (3 + cot^2(t))
Simplifying further, we get:
LHS = 1 - 2sin^2(t)
Thus, the left-hand side is equal to the right-hand side, proving the identity.

7. (1 - cos(@)) / (1 + cos(@)) = 2csc^2(@) - 2csc(@)cot(@) - 1
To prove this identity, let's simplify both sides:
LHS = (1 - cos(@)) / (1 + cos(@))
Multiplying the numerator and denominator by the conjugate of the denominator, we get:
LHS = (1 - cos(@)) * (1 - cos(@)) / (1 - cos^2(@))
= (1 - cos(@))^2 / sin^2(@)
= 1 - 2cos(@) + cos^2(@) / (1 - cos^2(@))
Applying the Pythagorean identity for sine, cos^2(@) = 1 - sin^2(@), we have:
LHS = 1 - 2cos(@) + 1 - sin^2(@) / (1 - (1 - sin^2(@)))
= 2 - 2cos(@) / sin^2(@)
= 2(1 - cos(@)) / sin^2(@)
= 2csc^2(@) - 2csc(@)cot(@)
Adding -1 to both sides, we get:
LHS - 1 = 2csc^2(@) - 2csc(@)cot(@) - 1
Thus, the left-hand side is equal to the right-hand side, proving the identity.

Therefore, identities 6 and 7 have been verified.