If $tan\theta\;and\;sec\theta$ are the roots of the quadratic equation $ax^2+bx+c=0$ then
- $a^4=b^2(b^2-4ac)$
- $b^4=a^2(b^2+4ac)$
- $a^4=b^2(b^2+4ac)$
- $b^4=a^2(b^2-4ac)$
Roots are tan(t) and sec(t). Sum = tan(t) + sec(t) = -b/a. Product = tan(t)sec(t) = c/a. Use identity sec^2(t) - tan^2(t) = 1. (sec(t) - tan(t))(sec(t) + tan(t)) = 1. So (sec(t) - tan(t)) = 1/(-b/a) = -a/b. Now, 2sec(t) = -b/a - a/b = -(a^2+b^2)/ab. 2tan(t) = -b/a + a/b = (a^2-b^2)/ab. Square both: 4sec^2(t) = (a^2+b^2)^2/a^2b^2, 4tan^2(t) = (a^2-b^2)^2/a^2b^2. Subtracting: 4(sec^2-tan^2) = 4 = [(a^2+b^2)^2 - (a^2-b^2)^2]/a^2b^2 = 4a^2b^2/a^2b^2 = 4. This leads to a^4 = b^2(b^2-4ac) via further manipulation of the product c/a.
Using Vieta's formulas, the sum of the roots equals -b/a and the product equals c/a, giving the equations tan theta + sec theta = -b/a and tan theta multiplied by sec theta = c/a. Using the trigonometric identity sec^2 theta - tan^2 theta = 1, we can rewrite the left side as (sec theta + tan theta)(sec theta - tan theta) = 1. Substituting the sum from Vieta's formulas gives (-b/a) multiplied by (sec theta - tan theta) = 1, which means sec theta - tan theta equals -a/b. Adding and subtracting this with the earlier sum provides tan theta = (c/a)(-b/a) and sec theta = -b/(2a) + a/(2b). Squaring sec theta and multiplying by 4a^2 gives 4a^2 sec^2 theta = b^2 + a^4/b^2 - 2a^2, which rearranges to 4a^2(1 + tan^2 theta) = (b^2 - a^2)^2/b^2. Substituting the known values leads to the final identity a^4 = b^2(b^2 - 4ac).