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Question: In \(\vartriangle ABC, (\dfrac{{{a^2}}}{{\sin A}} + \dfrac{{{b^2}}}{{\sin B}} + \dfrac{{{c^2}}}{{\si...

In △ABC,(a2sin⁡A+b2sin⁡B+c2sin⁡C).sin⁡A2sin⁡B2sin⁡C2\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) △2\dfrac{\vartriangle }{2}
D) △4\dfrac{\vartriangle }{4}

Explanation

Solution

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 (a2sin⁡A+b2sin⁡B+c2sin⁡C)(\dfrac{{{a^2}}}{{\sin A}} + \dfrac{{{b^2}}}{{\sin B}} + \dfrac{{{c^2}}}{{\sin C}})
And second part be sin⁡A2sin⁡B2sin⁡C2\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=a2sin⁡A=b2sin⁡B=c2sin⁡CR = \dfrac{a}{{2\sin A}} = \dfrac{b}{{2\sin B}} = \dfrac{c}{{2\sin C}}
From the above equation we write:
a=2Rsin⁡A2R\sin A
b=2Rsin⁡B2R\sin B
c=2Rsin⁡C2R\sin C
Put the values of a, b and c in the above equation:
a2sin⁡A=b2sin⁡B=c2sin⁡C=4R2(sin⁡A+sin⁡B+sin⁡C)\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
△=12bcsin⁡A=12acsin⁡B=12absin⁡C\vartriangle = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}ac\sin B = \dfrac{1}{2}ab\sin C
sin⁡A=2△bc,sin⁡B=2△ac,sin⁡C=2△ab\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
=4R2(2△bc+2△ac+2△ab)4{R^2}(\dfrac{{2\vartriangle }}{{bc}} + \dfrac{{2\vartriangle }}{{ac}} + \dfrac{{2\vartriangle }}{{ab}})
Taking 2△\vartriangle common and taking LCM
=8△R2abc(a+b+c)\dfrac{{8\vartriangle {R^2}}}{{abc}}(a + b + c)
=16△SR2abc\dfrac{{16\vartriangle S{R^2}}}{{abc}} [ since a+b+c=2S]……………… (a)
Now,
sin⁡A2sin⁡B2sin⁡C2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}
= (s−b)(s−c)bc(s−a)(s−c)ca(s−a)(s−b)ab\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
=(s−a)(s−b)(s−c)abc\dfrac{{(s - a)(s - b)(s - c)}}{{abc}}……………….. (b)
Now, we will find the product of (a) and (b)
16△SR2abc×(s−a)(s−b)(s−c)abc=16△3R2(abc)2\dfrac{{16\vartriangle S{R^2}}}{{abc}} \times \dfrac{{(s - a)(s - b)(s - c)}}{{abc}} = \dfrac{{16{\vartriangle ^3}{R^2}}}{{{{(abc)}^2}}}
=16△3(abc)2×(abc)216△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.