prove (tanθ + cotθ)^2 = sec^2 θ csc^2 θ

Answer 1

tanθ + cotθ = sinθ/cosθ + cosθ/sinθ = [sin²θ + cos²θ] / (sinθcosθ) =

1 / (sinθcosθ) = (1/sinθ)(1/cosθ) = secθcscθ

So, (tanθ + cotθ)² = (secθcscθ)² = sec²θ csc²θ

Answer 2

Square the binomial

tan² θ + 2 tan θ cot θ + cot² θ

sec² - 1 + 2 + csc² θ - 1 Pythagorean Identity

sec² θ+ csc² θ

QED