Question

In $\vartriangle ABC, (\dfrac{{{a^2}}}{{\sin A}} + \dfrac{{{b^2}}}{{\sin B}} + \dfrac{{{c^2}}}{{\sin C}}).\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$ simplifies to
A) 2$\vartriangle $
B) $\vartriangle $
C) $\dfrac{\vartriangle }{2}$
D) $\dfrac{\vartriangle }{4}$

Answer 1

Hint:In this question we will be using the application of trigonometry. Apply the formula and expand the given equation to solve it.

Complete step-by-step answer:
First we will divide our question in two parts as
Let the first part be $(\dfrac{{{a^2}}}{{\sin A}} + \dfrac{{{b^2}}}{{\sin B}} + \dfrac{{{c^2}}}{{\sin C}})$
And second part be $\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$
As we know that the radius of circumscribed circle is defined as:
$R = \dfrac{a}{{2\sin A}} = \dfrac{b}{{2\sin B}} = \dfrac{c}{{2\sin C}}$
From the above equation we write:
a=$2R\sin A$
b=$2R\sin B$
c=$2R\sin C$
Put the values of a, b and c in the above equation:
$\dfrac{{{a^2}}}{{\sin A}} = \dfrac{{{b^2}}}{{\sin B}} = \dfrac{{{c^2}}}{{\sin C}} = 4{R^2}(\sin A + \sin B + \sin C)$…………………. (1)
Now, we know that the area of the triangle is
$\vartriangle = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}ac\sin B = \dfrac{1}{2}ab\sin C$
$\sin A = \dfrac{{2\vartriangle }}{{bc}},\sin B = \dfrac{{2\vartriangle }}{{ac}},\sin C = \dfrac{{2\vartriangle }}{{ab}}$
Now, we will put the value of sinA , sinB and sinC in (1), we get
=$4{R^2}(\dfrac{{2\vartriangle }}{{bc}} + \dfrac{{2\vartriangle }}{{ac}} + \dfrac{{2\vartriangle }}{{ab}})$
Taking 2$\vartriangle $ common and taking LCM
=$\dfrac{{8\vartriangle {R^2}}}{{abc}}(a + b + c)$
=$\dfrac{{16\vartriangle S{R^2}}}{{abc}}$ [ since a+b+c=2S]……………… (a)
Now,
$\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$
= $$\sqrt {\dfrac{{(s - b)(s - c)}}{{bc}}} \sqrt {\dfrac{{(s - a)(s - c)}}{{ca}}} \sqrt {\dfrac{{(s - a)(s - b)}}{{ab}}} $$
On solving, we get
=$\dfrac{{(s - a)(s - b)(s - c)}}{{abc}}$……………….. (b)
Now, we will find the product of (a) and (b)
$\dfrac{{16\vartriangle S{R^2}}}{{abc}} \times \dfrac{{(s - a)(s - b)(s - c)}}{{abc}} = \dfrac{{16{\vartriangle ^3}{R^2}}}{{{{(abc)}^2}}}$
=$\dfrac{{16{\vartriangle ^3}}}{{{{(abc)}^2}}} \times \dfrac{{{{(abc)}^2}}}{{16{\vartriangle ^2}}}$
=$\vartriangle $

Note:
Individually solve the two parts of the question and then find the result, after that multiply the results obtained, use the formulas carefully.