Question

In a triangle $ABC$, $1 - \tan \left( {\dfrac{A}{2}} \right)\tan \left( {\dfrac{B}{2}} \right)$ is equal to
1) $\dfrac{{2c}}{{a + b + c}}$
2) $\dfrac{{2c}}{{a + b - c}}$
3) $\dfrac{{2b}}{{a + b + c}}$
4) None of these

Answer 1

Hint: -We will use the tangent rule for a triangle $ABC$, if the three sides of the triangle are given, that is,
$\tan \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} $
$\tan \left( {\dfrac{B}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{s\left( {s - b} \right)}}} $
$\tan \left( {\dfrac{C}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}}} $
Where, $s = \dfrac{{a + b + c}}{2}$
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 $ABC$, we get,
$\tan \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} $
$\tan \left( {\dfrac{B}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{s\left( {s - b} \right)}}} $
To find, $1 - \tan \left( {\dfrac{A}{2}} \right)\tan \left( {\dfrac{B}{2}} \right)$
Therefore, substituting these values in the required equation, we get,
$ = 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 - \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 - \sqrt {\dfrac{{\left( {s - c} \right)}}{s}.\dfrac{{\left( {s - c} \right)}}{s}} $
$ = 1 - \sqrt {\dfrac{{{{\left( {s - c} \right)}^2}}}{{{s^2}}}} $
Therefore, taking the square root, we get,
$ = 1 - \dfrac{{\left( {s - c} \right)}}{s}$
Now, dividing the numerator by the denominator, we get,
$ = 1 - \left( {1 - \dfrac{c}{s}} \right)$
Now, opening the brackets, we get,
$ = 1 - 1 + \dfrac{c}{s}$
$ = \dfrac{c}{s}$
We know, $s = \dfrac{{a + b + c}}{2}$.
Therefore, substituting this value, we get,
$ = \dfrac{c}{{\dfrac{{a + b + c}}{2}}}$
We can also write it as,
$ = \dfrac{{2c}}{{a + b + c}}$
Therefore, $1 - \tan \left( {\dfrac{A}{2}} \right)\tan \left( {\dfrac{B}{2}} \right) = \dfrac{{2c}}{{a + b + c}}$, 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,
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$
This is the sine rule.
And, $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
$\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$
$\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
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.