Question

In \(\triangle ABC\), with usual notations, the value of \(\text{cosec}\left(\frac{A}{2}\right)\text{cosec}\left(\frac{B}{2}\right)\text{cosec}\left(\frac{C}{2}\right)\) is:

• A. \(\frac{[A\triangle ABC]^{2}}{abcs}\)
• B. \(\frac{abcs}{[A\triangle ABC]^{2}}\)
• C. \(\frac{[A\triangle ABC]}{abcs}\)
• D. \(\frac{abcs}{[A\triangle ABC]}\)

Hint:

Use cosec(A/2) = sqrt(bc/((s-b)(s-c))) and multiply the three results.

Answer 1

The Correct Option is B

Step 1: Half Angle Formula
\(\sin\dfrac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{bc}}\), so \(\text{cosec}\dfrac{A}{2}=\sqrt{\dfrac{bc}{(s-b)(s-c)}}\). Similar results hold for \(B\) and \(C\).

Step 2: Multiply
\(\text{cosec}\dfrac{A}{2}\,\text{cosec}\dfrac{B}{2}\,\text{cosec}\dfrac{C}{2}=\sqrt{\dfrac{a^{2}b^{2}c^{2}}{(s-a)^{2}(s-b)^{2}(s-c)^{2}}}=\dfrac{abc}{(s-a)(s-b)(s-c)}\).

Step 3: Use Heron's Formula
\(\Delta^{2}=s(s-a)(s-b)(s-c)\), so \((s-a)(s-b)(s-c)=\dfrac{\Delta^{2}}{s}\).

Step 4: Simplify
The product equals \(\dfrac{abcs}{\Delta^{2}}\), option (B).

Final Answer:
The product equals \(\frac{abcs}{[A\triangle ABC]^{2}}\). \[ \boxed{\text{(B) }\dfrac{abcs}{[A\triangle ABC]^{2}}} \]