How do you prove sec^2((pi/2)-x)-1=cot^2x?

1 Answer
Nov 21, 2015

We will be using the following properties:

  1. sec(x) = 1/cos(x) (definition of secant)
  2. cot(x) = cos(x)/sin(x) (definition of cotangent)
  3. cos(-x) = cos(x) (cosine is even)
  4. cos(x - pi/2) = sin(x)
  5. sin^2(x) + cos^2(x) = 1

By 1, we have

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

By 3 and 4,

cos^2(pi/2-x) = cos^2(-(x - pi/2)) = cos^2(x-pi/2) = sin^2(x)

=> 1/(cos^2(pi/2-x))-1 = 1/sin^2(x)-1

By 5,

1/sin^2(x) = (sin^2(x) + cos^2(x))/sin^2(x) = sin^2(x)/sin^2(x) + cos^2(x)/sin^2(x)

Thus, by 2,

1/sin^2(x) = 1 + cot^2(x)

Substituting back then gives us

1/(cos^2(pi/2-x))-1 = (1 + cot^2(x))-1

:. sec^2(pi/2-x)-1 = cot^2(x) " Q.E.D."