Question

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.

Answer 1

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

remember that the co- in <u>co</u>sine, <u>co</u>tan, <u>co</u>sec means the <u>co</u>mplementary angle.

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

Answer 2

oops. That is

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