Question

In a $\triangle ABC$, $\frac{2(r_1 + r_3)}{a c (1 + \cos B)} =$

• A. $\frac{\Delta}{b}$
• B. $\frac{b}{\Delta}$
• C. $\frac{a + b + c}{2\Delta}$
• D. $\frac{a + b - c}{2\Delta}$

Hint:

Use exradius formulas and half-angle identities to simplify expressions involving $r_1, r_3$. Relate to $\Delta$ and side lengths via sine and cosine rules.

Answer 1

The Correct Option is B

Use the exradius formulas: $r_1 = 4R \sin\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right) \cos\left(\frac{C}{2}\right)$, $r_3 = 4R \sin\left(\frac{C}{2}\right) \cos\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right)$. Thus: \[ r_1 + r_3 = 4R \cos\left(\frac{B}{2}\right) \left[ \sin\left(\frac{A}{2}\right) \cos\left(\frac{C}{2}\right) + \cos\left(\frac{A}{2}\right) \sin\left(\frac{C}{2}\right) \right] = 4R \cos\left(\frac{B}{2}\right) \sin\left(\frac{A + C}{2}\right) \] Since $A + B + C = 180^\circ$, $\frac{A + C}{2} = 90^\circ - \frac{B}{2}$, so $\sin\left(\frac{A + C}{2}\right) = \cos\left(\frac{B}{2}\right)$. Hence: \[ r_1 + r_3 = 4R \cos^2\left(\frac{B}{2}\right) = 2R (1 + \cos B) \quad \text{(since } 2 \cos^2\left(\frac{B}{2}\right) = 1 + \cos B\text{)} \] Compute the denominator: $a = 2R \sin A$, $c = 2R \sin C$, so: \[ a c = (2R \sin A)(2R \sin C) = 4R^2 \sin A \sin C \] \[ \frac{2 (r_1 + r_3)}{a c (1 + \cos B)} = \frac{2 \cdot 2R (1 + \cos B)}{4R^2 \sin A \sin C \cdot (1 + \cos B)} = \frac{4R}{4R^2 \sin A \sin C} = \frac{1}{R \sin A \sin C} \] Since $b = 2R \sin B$, $R = \frac{b}{2 \sin B}$. The area $\Delta = \frac{1}{2} a c \sin B = \frac{1}{2} (2R \sin A)(2R \sin C) \sin B = 2R^2 \sin A \sin B \sin C$. Thus: \[ \frac{1}{R \sin A \sin C} = \frac{2 \sin B}{2R \sin A \sin B \sin C} = \frac{2 \sin B}{\frac{a c \sin B}{2}} = \frac{4 \sin B}{a c \sin B} = \frac{b}{\Delta} \] Option (2) is correct. Options (1), (3), and (4) do not match the derived expression.