If sinθ − cosθ = 1/√5, then tanθ + cotθ equals:

Squaring gives 1 − 2sinθ·cosθ = 1/5, so sinθ·cosθ = 2/5. Since tanθ + cotθ = (si$n^2$θ + co$s^2$θ)/(sinθ·cosθ) = 1/(sinθ·cosθ), the value is 5/2. Reporting the product itself gives 2/5, forgetting to halve after the subtraction gives 5/4, and using 1 + 2sinθ·cosθ — the sign that belongs to the sum, not the difference — gives −5/2.