Did you notice that pi/2-x and x are complementary angles (add up to 90ยบ)?
so we know that cos(pi/2-x) = sinx
I will use that ...
LS
= sec^2 (pi/2 - x) -1
= 1/cos^2 (pi/2-x) - 1
= 1/[cos(pi/2-x)cos(pi/2-x)] - 1
= 1/[sinxsinx] - 1
= 1/sin^2x - 1
= (1 - sin^2x)/sin^2x
= cos^2 x/sin^2 x
= cot^2 x
= RS
1) verify the identity
sec^2 (pi/2 - x) -1 = cot^2 x
I think that we can replace sec with csc for the cofunction formulas since there is pi/2. Since there is squared and a 1 I think we could use one of the pythagorean identites but I am not sure which one.
1 answer