Trig verifying identities

I am having trouble with this problem.

sec^2(pi/2-x)-1= cot ^2x

I got :

By cofunction identity sec(90 degrees - x) = csc x

secx csc-1 = cot^2x

Then split sec x and csc-1 into two fractions and multiplied both numerator and denominators by csc and got:

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

Then by Pythagorean identity I got:

sec x cot^2= cot^2
I got stuck here because I could not figure out how to get rid of the sec x so I just put down false; the equations are not equal.

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asked by Hutch
  1. sec(pi/2 - x) = csc x
    sec^2(x) - 1 = cot^2(x)

    remember that the co- in cosine, cotan, cosec means the complementary angle.

    cos(x) = sin(pi/2 - x)
    and so on

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    posted by Steve
  2. oops. That is

    csc^2(x) - 1 = cot^2(x)

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    posted by Steve

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