Question

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

Answer 1

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`

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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."`