Question

In a triangle ABC, \( (r_2 + r_3)\sec^2\left(\frac{A}{2}\right) = \)
(Note: Based on the answer key and standard identities, the function is interpreted as \( \sec^2 \)).

• A. \( 4R \cot\left(\frac{A}{2}\right) \)
• B. \( 2R \cot^2\left(\frac{A}{2}\right) \)
• C. \( 4R \)
• D. \( 2R \tan\left(\frac{A}{2}\right) \)

Hint:

The identity \( r_2 + r_3 = 4R \cos^2(A/2) \) is very useful.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We simplify the expression involving ex-radii and trigonometric functions.
Step 2: Key Formula or Approach:

\( r_2 = 4R \cos\frac{A}{2} \sin\frac{B}{2} \cos\frac{C}{2} \)
\( r_3 = 4R \cos\frac{A}{2} \cos\frac{B}{2} \sin\frac{C}{2} \)
Step 3: Detailed Explanation:

Sum of \( r_2 \) and \( r_3 \): \[ r_2 + r_3 = 4R \cos\frac{A}{2} \left( \sin\frac{B}{2} \cos\frac{C}{2} + \cos\frac{B}{2} \sin\frac{C}{2} \right) \] Using \( \sin(x+y) \): \[ r_2 + r_3 = 4R \cos\frac{A}{2} \sin\left(\frac{B+C}{2}\right) \] Since \( A+B+C = \pi \), \( \frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2} \), so \( \sin\left(\frac{B+C}{2}\right) = \cos\frac{A}{2} \). \[ r_2 + r_3 = 4R \cos^2\frac{A}{2} \] Now multiply by \( \sec^2\left(\frac{A}{2}\right) \): \[ (r_2 + r_3) \sec^2\left(\frac{A}{2}\right) = 4R \cos^2\frac{A}{2} \cdot \frac{1}{\cos^2\frac{A}{2}} = 4R \]
Step 4: Final Answer:

The value is \( 4R \).