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
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.
2 answers
oops. That is
csc^2(x) - 1 = cot^2(x)
csc^2(x) - 1 = cot^2(x)