Question

In \(△ABC\) with usual notations, if \(1+tan(\frac{A}{2})tan(\frac{B}{2}) = \frac{k}{s}\) (where \(s\) is the semi-perimeter), then the value of \(k\) is...

• A. \(2\)
• B. \(a+b-c\)
• C. \(a+b\)
• D. \(s-c\)

Hint:

Use tan(A/2)tan(B/2) = (s-c)/s.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a triangle, \(\tan\dfrac A2\tan\dfrac B2=\dfrac{s-c}{s}\) (half-angle formulae).

Step 2: Add 1:
\[ 1+\tan\tfrac A2\tan\tfrac B2=1+\frac{s-c}{s}=\frac{2s-c}{s} \]

Step 3: Use the perimeter:
\(2s=a+b+c\), so \(2s-c=a+b\). Hence
\[ 1+\tan\tfrac A2\tan\tfrac B2=\frac{a+b}{s} \]

Step 4: Compare with k/s:
So \(k=a+b\), option (C). Option (B) \(a+b-c\) is \(2(s-c)\) and option (D) \(s-c\) would give \(\frac{s-c}{s}\) without the 1.

Final Answer:
k equals a + b. \[ \boxed{k=a+b} \]