Animated Solution for Mathematics - Trigonometry: In a triangle $PQR,\angle R=\pi /2$. If $\tan (P/2)$ and $\tan (Q/2)$ are the roots of the equation $a{x}^2+bx+c=0(a\ne 0)$ then.

In a right-angled triangle $PQR$ where $\angle R=\frac{\pi }{2}$, the sum of the interior angles is constrained by $P+Q+R=\pi$. Substituting the known value of $\angle R$, we find that $P+Q=\frac{\pi }{2}$.

Since the problem involves the half-angles $\frac{P}{2}$ and $\frac{Q}{2}$, we divide our constraint by two to obtain:

This value, $\frac{\pi }{4}$, serves as our anchor for the subsequent trigonometric calculations.

Connecting Roots to Coefficients

Consider the quadratic equation $a{x}^2+bx+c=0$ with roots $\tan (\frac{P}{2})$ and $\tan (\frac{Q}{2})$. According to Vieta's formulas, the sum and product of these roots are related to the coefficients $a,b,$ and $c$ as follows:

These equations successfully translate the geometric properties of the triangle into the algebraic language of coefficients.

The Final Elegance

To synthesize these results, we utilize the tangent addition formula:

Setting $A=\frac{P}{2}$ and $B=\frac{Q}{2}$, and knowing that $\tan (\frac{\pi }{4})=1$, we substitute our Vieta's expressions into the formula:

To simplify, we multiply the numerator and denominator by $a$ to clear the fractions:

Cross-multiplying yields $a-c=-b$. Rearranging the terms, we arrive at the final relation: