A quadratic equation whose roots are $\displaystyle \tan 22\frac{1}{2}^{0}$ and $\displaystyle \cot 22 \frac{1}{2}^{0}$ is
- $x^{2}-2\sqrt{2}x+1=0$
- $x^{2_{-}}\sqrt{x}+1=0$
- $x^{2}+2\sqrt{2}x-1=0$
- $x^{2}-2\sqrt{2}-1=0$
Let x1 = tan(22.5) and x2 = cot(22.5). Since cot(22.5) = 1/tan(22.5), the product of roots is 1. The sum of roots is tan(22.5) + cot(22.5) = (sin^2(22.5) + cos^2(22.5)) / (sin(22.5)cos(22.5)) = 1 / (0.5 * sin(45)) = 1 / (0.5 * 1/sqrt(2)) = 2*sqrt(2). The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 - 2*sqrt(2)x + 1 = 0.
Let the roots of the quadratic equation be r1 = tan(22.5 degrees) and r2 = cot(22.5 degrees). The sum of the roots is tan(22.5 degrees) + cot(22.5 degrees) = tan(22.5 degrees) + 1/tan(22.5 degrees). Using the double angle identity for cotangent, cot(22.5 degrees) = cot(45 degrees / 2) = cot(45 degrees) + cosec(45 degrees) = 1 + sqrt(2), and tan(22.5 degrees) = 1 / (1 + sqrt(2)) = sqrt(2) - 1. The sum is (sqrt(2) - 1) + (1 + sqrt(2)) = 2*sqrt(2), and the product is 1 because they are reciprocals. The required quadratic equation is x^2 - (Sum)x + (Product) = 0, so we get x^2 - 2*sqrt(2)x + 1 = 0.