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Question: In a triangle \( ABC \) if \( \cot \dfrac{A}{2}.\cot \dfrac{B}{2} = c,\cot \dfrac{B}{2}.\cot \dfrac{...

In a triangle ABCABC if cot⁡A2.cot⁡B2=c,cot⁡B2.cot⁡C2=a\cot \dfrac{A}{2}.\cot \dfrac{B}{2} = c,\cot \dfrac{B}{2}.\cot \dfrac{C}{2} = a and cot⁡C2.cot⁡A2=b\cot \dfrac{C}{2}.\cot \dfrac{A}{2} = b then 1s−a+1s−b+1s−c=?\dfrac{1}{{s - a}} + \dfrac{1}{{s - b}} + \dfrac{1}{{s - c}} = ?
A) −1- 1
B) 00
C) 11
D) 22

Explanation

Solution

Hint : To solve this question we use trigonometric ratios of half angle and then substitute the cot values . Thus, we will have three equations. When we compare these three solutions we will get the solution. We also use heron’s formula in this question.

Complete step-by-step answer :
According to the trigonometric ratios of half angle, we know that
tan⁡A2=(s−b)(s−c)s(s−a)\tan \dfrac{A}{2} = \sqrt {\dfrac{{(s - b)(s - c)}}{{s(s - a)}}}
Then,
cot⁡A2=s(s−a)(s−b)(s−c)\cot \dfrac{A}{2} = \sqrt {\dfrac{{s(s - a)}}{{(s - b)(s - c)}}}
Now using this formula , we will solve this problem.
It is given in the question that
cot⁡A2.cot⁡B2=c s(s−a)(s−b)(s−c).s(s−b)(s−a)(s−c)=c s(s−a)(s−b)(s−c).s(s−b)(s−a)(s−c)=c ss−c=c−(i)   \cot \dfrac{A}{2}.\cot \dfrac{B}{2} = c \\\ \sqrt {\dfrac{{s(s - a)}}{{(s - b)(s - c)}}} .\sqrt {\dfrac{{s(s - b)}}{{(s - a)(s - c)}}} = c \\\ \sqrt {\dfrac{{s(s - {a})}}{{(s - {b})(s - c)}}} .\sqrt {\dfrac{{s(s - {b})}}{{(s - {a})(s - c)}}} = c \\\ \dfrac{s}{{s - c}} = c - (i) \;
Now,
cot⁡B2.cot⁡C2=a s(s−b)(s−a)(s−c).s(s−c)(s−a)(s−b)=a s(s−b)(s−a)(s−c).s(s−c)(s−a)(s−b)=a ss−a=a−(ii)   \cot \dfrac{B}{2}.\cot \dfrac{C}{2} = a \\\ \sqrt {\dfrac{{s(s - b)}}{{(s - a)(s - c)}}} .\sqrt {\dfrac{{s(s - c)}}{{(s - a)(s - b)}}} = a \\\ \sqrt {\dfrac{{s(s - {b})}}{{(s - a)(s - {c})}}} .\sqrt {\dfrac{{s(s - {c})}}{{(s - a)(s - {b})}}} = a \\\ \dfrac{s}{{s - a}} = a - (ii) \;
Now,
cot⁡C2.cot⁡A2=b s(s−c)(s−a)(s−b).s(s−a)(s−b)(s−c)=b s(s−c)(s−a)(s−b).s(s−a)(s−b)(s−c)=b ss−b=b−(iii)   \cot \dfrac{C}{2}.\cot \dfrac{A}{2} = b \\\ \sqrt {\dfrac{{s(s - c)}}{{(s - a)(s - b)}}} .\sqrt {\dfrac{{s(s - a)}}{{(s - b)(s - c)}}} = b \\\ \sqrt {\dfrac{{s(s - {c})}}{{(s - {a})(s - b)}}} .\sqrt {\dfrac{{s(s - {a})}}{{(s - b)(s - {c})}}} = b \\\ \dfrac{s}{{s - b}} = b - (iii) \;
Now, let us add the three equations
ss−c+ss−a+ss−b=c+a+b s(1s−c+1s−a+1s−b)=c+a+b 1s−c+1s−a+1s−b=a+b+cs   \dfrac{s}{{s - c}} + \dfrac{s}{{s - a}} + \dfrac{s}{{s - b}} = c + a + b \\\ s\left( {\dfrac{1}{{s - c}} + \dfrac{1}{{s - a}} + \dfrac{1}{{s - b}}} \right) = c + a + b \\\ \dfrac{1}{{s - c}} + \dfrac{1}{{s - a}} + \dfrac{1}{{s - b}} = \dfrac{{a + b + c}}{s} \;
According to heron’s formula we know that, s=a+b+c2s = \dfrac{{a + b + c}}{2}
1s−c+1s−a+1s−b=2ss 1s−c+1s−a+1s−b=2   \dfrac{1}{{s - c}} + \dfrac{1}{{s - a}} + \dfrac{1}{{s - b}} = \dfrac{{2{s}}}{{{s}}} \\\ \dfrac{1}{{s - c}} + \dfrac{1}{{s - a}} + \dfrac{1}{{s - b}} = 2 \;
Hence, the correct option is (D) which is 2.
So, the correct answer is “OPTION D”.

Note : The sine and cosine formula is derived using the basic formulas of triangle which depicts the relation between sides and angles of triangle. Then these formulas are further used to derive the trigonometric ratios of half angles.