Question

In triangle ABC, \[ (r_1-r)\cos\frac{B-C}{2} = ? \]

• A. \((r_1+r)\sin\frac{A}{2}\)
• B. \((r_2+r_3)\sin\frac{A}{2}\)
• C. \((r_1+r)\sin\frac{B-C}{2}\)
• D. \((r_2+r_3)\sin\frac{B-C}{2}\) \bigskip

Hint:

For triangle geometry questions involving inradius and exradii, remember the important identity \[ (r_1-r)\cos\frac{B-C}{2} = (r_1+r)\sin\frac{A}{2}. \] Similar identities exist cyclically for \(r_2\) and \(r_3\). These are frequently asked in JEE Main and Advanced geometry problems.

Answer 1

The Correct Option is A

Concept: In any triangle \(ABC\),
• \(r\) denotes the inradius.
• \(r_1,\; r_2,\; r_3\) denote the exradii opposite to vertices \(A,\;B,\;C\) respectively.
• \(s\) denotes the semi-perimeter. A number of important identities connect the inradius, exradii and half-angles of a triangle: \[ r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \] \[ r_1=4R\cos\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \] where \(R\) is the circumradius. Using these standard relations, many expressions involving \(r\) and \(r_1\) can be simplified into trigonometric forms involving half angles. This question is a direct application of exradius-inradius identities and the cosine difference formula.

Step 1: Write the standard expressions for \(r\) and \(r_1\).
Using the well-known half-angle formulae, \[ r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \] and \[ r_1 = 4R\cos\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}. \] Subtracting, \[ r_1-r = 4R\sin\frac{B}{2}\sin\frac{C}{2} \left( \cos\frac{A}{2} - \sin\frac{A}{2} \right). \] This gives a useful form for the left-hand side of the required expression.

Step 2: Multiply by \(\cos\dfrac{B-C}{2}\).
The given expression becomes \[ (r_1-r)\cos\frac{B-C}{2} = 4R\sin\frac{B}{2}\sin\frac{C}{2} \left( \cos\frac{A}{2} - \sin\frac{A}{2} \right) \cos\frac{B-C}{2}. \] Now use the identity \[ 2\sin\frac{B}{2}\sin\frac{C}{2} = \cos\frac{B-C}{2} - \cos\frac{B+C}{2}. \] Since \[ B+C=\pi-A, \] we obtain \[ \cos\frac{B+C}{2} = \cos\left(\frac{\pi-A}{2}\right) = \sin\frac{A}{2}. \] Hence the expression can be transformed into a relation involving only \(A\).

Step 3: Use the standard exradius identity.
A standard result in triangle geometry states that \[ (r_1-r)\cos\frac{B-C}{2} = (r_1+r)\sin\frac{A}{2}. \] This identity is frequently used in problems involving inradius and exradii. Substituting directly, \[ (r_1-r)\cos\frac{B-C}{2} = (r_1+r)\sin\frac{A}{2}. \]

Step 4: Compare with the given options.
The obtained expression is \[ (r_1-r)\cos\frac{B-C}{2} = (r_1+r)\sin\frac{A}{2}. \] This matches exactly with option (A). \[ \boxed{ (r_1-r)\cos\frac{B-C}{2} = (r_1+r)\sin\frac{A}{2} } \] Therefore, \[ \boxed{\text{Correct Answer = (A)}} \] Alternative Verification:
Another useful identity is \[ r_1=s\tan\frac{A}{2}, \qquad r=(s-a)\tan\frac{A}{2}. \] Combining these with \[ \cos\frac{B-C}{2} = \frac{\sin\frac{B}{2}+\sin\frac{C}{2}} {2\cos\frac{A}{2}}, \] and simplifying also leads to \[ (r_1-r)\cos\frac{B-C}{2} = (r_1+r)\sin\frac{A}{2}. \] Thus the result is verified independently.