Find the intervals on which f(x) = sec(x)csc(x), 0 <= x <= 2pi, is increasing.

Answer 1

f(x) = (secx)(cscx) = [cosx sinx]^-1. f(x) is undefined at x = 0, π/2, π, 3π/2, 2π

f'(x) = -1[cosx sinx]^-2[cos²x - sin²x] = -cos(2x) / (cosx sinx)²

f'(x) = 0 when 2x = π/2, 3π/2, 5π/2, 7π/2, .... So, x = π/4, 3π/4, 5π/4, 7π/4...

On 0 < x < π/4, f'(x) < 0. So, f is decreasing

On π/4 < x < π/2, f'(x) > 0. So, f is increasing

On π/2 < x < 3π/4, f'(x) > 0. So, f is increasing

On 3π/4 < x < π, f'(x) < 0. So, f is decreasing

On π < x < 5π/4, f'(x) < 0. So, f is decreasing

On 5π/4 < x < 3π/2, f'(x) > 0. So, f is increasing

On 3π/2 < x < 7π/4, f'(x) > 0. So, f is increasing

On 7π/4 < x < 2π, f'(x) < 0. So, f is decreasing

f(x) is increasing on (π/4. π/2) ∪ (π/2, 3π/4) ∪ (5π/4, 3π/2) ∪ (3π/2, 7π/4)