In $\triangle PQR, \,\angle R = \dfrac { \pi }{ 2 } $. If $\tan \left(\dfrac{P}{2}\right)$ and $\tan \left(\dfrac{Q}{2}\right)$ are the roots of the equation $ax^2 + bx + c= 0$, then which one of the following is correct?
- $a = b + c$
- $b = c + a $
- $c= a+ b$
- $ b = c$
In a right triangle, P + Q = 90 degrees, so P/2 + Q/2 = 45 degrees. Thus, tan(P/2 + Q/2) = tan(45) = 1. Using the tangent sum formula, (tan(P/2) + tan(Q/2)) / (1 - tan(P/2)tan(Q/2)) = 1. Let roots be x1, x2. Then (x1+x2) / (1-x1x2) = 1. From the quadratic, x1+x2 = -b/a and x1x2 = c/a. Substituting gives (-b/a) / (1 - c/a) = 1, which simplifies to -b / (a-c) = 1, so -b = a - c, or c = a + b.
In triangle PQR, angle P + angle Q = pi/2 because angle R is pi/2. Let tan(P/2) = p and tan(Q/2) = q, meaning p + q = -b/a and pq = c/a from the given quadratic. Using the double angle formula, tan P = 2p / (1 - p squared) and tan Q = 2q / (1 - q squared). Since P and Q are complementary, tan P tan Q = 1, leading to 4pq = (1 - p squared)(1 - q squared). Expanding this gives 4pq = 1 - (p squared + q squared) + p squared q squared. Substituting p squared + q squared = (p + q) squared - 2pq results in a complex relation, but we can use the sine rule instead. We know sin P = 2pq / (1 + p squared)(1 + q squared) and cos P = (1 - p squared)(1 - q squared) / (1 + p squared)(1 + q squared). Since sin P + cos P = 1 is not generally true, we test the options by finding the tangent of the sum of the half angles. Using the formula tan(P/2 + Q/2) = (p + q) / (1 - pq) and knowing P/2 + Q/2 = pi/4, we have (p + q) / (1 - pq) = tan(pi/4) = 1. This simplifies to p + q = 1 - pq, or p + q + pq = 1. Substituting the Vieta's formulas gives -b/a + c/a = 1, which means c - b = a. Therefore, c = a + b.