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Question: In a triangle \(ABC\), \(1 - \tan \left( {\dfrac{A}{2}} \right)\tan \left( {\dfrac{B}{2}} \right)\) ...

In a triangle ABCABC, 1−tan⁡(A2)tan⁡(B2)1 - \tan \left( {\dfrac{A}{2}} \right)\tan \left( {\dfrac{B}{2}} \right) is equal to

  1. 2ca+b+c\dfrac{{2c}}{{a + b + c}}
  2. 2ca+b−c\dfrac{{2c}}{{a + b - c}}
  3. 2ba+b+c\dfrac{{2b}}{{a + b + c}}
  4. None of these
Explanation

Solution

Hint : -We will use the tangent rule for a triangle ABCABC, if the three sides of the triangle are given, that is,
tan⁡(A2)=(s−b)(s−c)s(s−a)\tan \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}}
tan⁡(B2)=(s−a)(s−c)s(s−b)\tan \left( {\dfrac{B}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{s\left( {s - b} \right)}}}
tan⁡(C2)=(s−a)(s−b)s(s−c)\tan \left( {\dfrac{C}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}}}
Where, s=a+b+c2s = \dfrac{{a + b + c}}{2}
After finding the values of the required terms, then we will substitute the terms required and perform the required operations to find the required solution.

Complete step-by-step answer :
We know, by the tangent rule of a triangle ABCABC, we get,
tan⁡(A2)=(s−b)(s−c)s(s−a)\tan \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}}
tan⁡(B2)=(s−a)(s−c)s(s−b)\tan \left( {\dfrac{B}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{s\left( {s - b} \right)}}}
To find, 1−tan⁡(A2)tan⁡(B2)1 - \tan \left( {\dfrac{A}{2}} \right)\tan \left( {\dfrac{B}{2}} \right)
Therefore, substituting these values in the required equation, we get,
=1−(s−b)(s−c)s(s−a)(s−a)(s−c)s(s−b)= 1 - \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{s\left( {s - b} \right)}}}
=1−(s−b)(s−c)s(s−a).(s−a)(s−c)s(s−b)= 1 - \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}.\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{s\left( {s - b} \right)}}}
Now, cancelling the similar terms, we get,
=1−(s−c)s.(s−c)s= 1 - \sqrt {\dfrac{{\left( {s - c} \right)}}{s}.\dfrac{{\left( {s - c} \right)}}{s}}
=1−(s−c)2s2= 1 - \sqrt {\dfrac{{{{\left( {s - c} \right)}^2}}}{{{s^2}}}}
Therefore, taking the square root, we get,
=1−(s−c)s= 1 - \dfrac{{\left( {s - c} \right)}}{s}
Now, dividing the numerator by the denominator, we get,
=1−(1−cs)= 1 - \left( {1 - \dfrac{c}{s}} \right)
Now, opening the brackets, we get,
=1−1+cs= 1 - 1 + \dfrac{c}{s}
=cs= \dfrac{c}{s}
We know, s=a+b+c2s = \dfrac{{a + b + c}}{2}.
Therefore, substituting this value, we get,
=ca+b+c2= \dfrac{c}{{\dfrac{{a + b + c}}{2}}}
We can also write it as,
=2ca+b+c= \dfrac{{2c}}{{a + b + c}}
Therefore, 1−tan⁡(A2)tan⁡(B2)=2ca+b+c1 - \tan \left( {\dfrac{A}{2}} \right)\tan \left( {\dfrac{B}{2}} \right) = \dfrac{{2c}}{{a + b + c}}, the correct option is 1.
So, the correct answer is “Option 1”.

Note : Here, in this question we used the tangent rule, but along with the tangent rules, there are also other rules like the sine and cosine rule, that says,
asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}
This is the sine rule.
And, cos⁡A=b2+c2−a22bc\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}
cos⁡B=a2+c2−b22ac\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}
cos⁡C=a2+b2−c22ab\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}
This is the cosine rule.
So, on the basis of the question and as per requirement we can use these rules to solve the problem.