Key Concepts and Formulas

• Angles in a Triangle: The sum of the angles in any triangle is $\pi$ radians (180 degrees). In a right-angled triangle, the two acute angles sum to $\pi/2$ radians (90 degrees).
• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $x_1$ and $x_2$, the sum of the roots is $x_1 + x_2 = -\frac{b}{a}$ and the product of the roots is $x_1 x_2 = \frac{c}{a}$.
• Tangent Addition Formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

Step-by-Step Solution

Step 1: Establish the relationship between angles P and Q.

Since triangle $PQR$ is a right-angled triangle with $\angle R = \frac{\pi}{2}$, we have $\angle P + \angle Q + \angle R = \pi$. Substituting $\angle R = \frac{\pi}{2}$, we get $\angle P + \angle Q + \frac{\pi}{2} = \pi$, which simplifies to $\angle P + \angle Q = \frac{\pi}{2}$.

Why? This establishes the fundamental relationship between the two acute angles in the right triangle, which is crucial for relating the trigonometric functions of their halves.

Step 2: Determine the sum of half-angles.

Dividing the equation $\angle P + \angle Q = \frac{\pi}{2}$ by 2, we get $\frac{\angle P}{2} + \frac{\angle Q}{2} = \frac{\pi}{4}$.

Why? This gives us the value of the sum of the arguments of the tangent functions, which will be used in the tangent addition formula.

Step 3: Apply Vieta's formulas to the quadratic equation.

Given that $\tan\left(\frac{P}{2}\right)$ and $\tan\left(\frac{Q}{2}\right)$ are the roots of the equation $ax^2 + bx + c = 0$, we can use Vieta's formulas to express the sum and product of the roots in terms of the coefficients $a$, $b$, and $c$. Let $x_1 = \tan\left(\frac{P}{2}\right)$ and $x_2 = \tan\left(\frac{Q}{2}\right)$. Then:

• $x_1 + x_2 = \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = -\frac{b}{a}$
• $x_1 x_2 = \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right) = \frac{c}{a}$

Why? Vieta's formulas provide the link between the roots (expressed as tangent functions) and the coefficients of the quadratic equation.

Step 4: Use the tangent addition formula.

We have $\tan\left(\frac{P}{2} + \frac{Q}{2}\right) = \frac{\tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right)}{1 - \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right)}$.

Why? This formula allows us to relate the sum and product of the tangents of the half-angles to the tangent of their sum.

Step 5: Substitute known values into the tangent addition formula.

Since $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$, we have $\tan\left(\frac{P}{2} + \frac{Q}{2}\right) = \tan\left(\frac{\pi}{4}\right) = 1$. Substituting the values from Vieta's formulas into the tangent addition formula: $1 = \frac{-\frac{b}{a}}{1 - \frac{c}{a}}$

Why? This substitution combines all the information to form a single equation relating $a$, $b$, and $c$.

Step 6: Simplify and solve for the relationship between a, b, and c.

Simplifying the equation, we get: $1 = \frac{-\frac{b}{a}}{\frac{a - c}{a}}$ $1 = \frac{-b}{a - c}$ $a - c = -b$ $a + b = c$ Therefore, $c = a + b$.

Why? This algebraic manipulation isolates the relationship between the coefficients, leading to the solution.

Common Mistakes & Tips

• Ensure you are using the correct trigonometric identity (tangent addition formula).
• Be careful with algebraic manipulations, particularly when simplifying fractions and handling negative signs.
• Remember the relationships between angles in a right-angled triangle.

Summary

By using the properties of right-angled triangles, Vieta's formulas, and the tangent addition formula, we were able to establish a relationship between the coefficients of the quadratic equation. The final relationship is $c = a + b$.

Final Answer

The final answer is \boxed{c = a + b}, which corresponds to option (B).