Question

Prove that $\dfrac{{{a^2}\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{{b^2}\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{{c^2}\sin (A - B)}}{{\sin A + \sin B}} = 0$

Answer 1

Here in this question properties of triangle in trigonometric functions will get used. Radius of circumscribed circle also called as derivation of sine rule is given by: -$R = \dfrac{a}{{2\sin A}} = \dfrac{b}{{2\sin B}} = \dfrac{c}{{2\sin C}}$
$a = R2\sin A$, $b = R2\sin B$,$c = R2\sin C$

Complete step-by-step answer:
L.H.S=$\dfrac{{{a^2}\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{{b^2}\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{{c^2}\sin (A - B)}}{{\sin A + \sin B}}$
First of all we will simplify the numerator by putting values of a, b, and c from the formula of derivation of sine rule.
$a = R2\sin A$, $b = R2\sin B$,$c = R2\sin C$ $\Rightarrow \dfrac{{{{(R2\sin A)}^2}\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{{{(R2\sin B)}^2}\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{{{(R2\sin C)}^2}\sin (A - B)}}{{\sin A + \sin B}}$
Now we will solve the square terms in the bracket to simplify further.
$\Rightarrow \dfrac{{({R^2}4{{\sin }^2}A)\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{({R^2}4{{\sin }^2}B)\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{({R^2}4{{\sin }^2}C)\sin (A - B)}}{{\sin A + \sin B}}$
Now we will apply property of triangle in trigonometric functions i.e. $\sin A = \sin (\pi - (B + C))$ \Rightarrow \dfrac{{({R^2}4\sin A)\sin A\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{({R^2}4{{\sin }^2}B)\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{({R^2}4{{\sin }^2}C)\sin (A - B)}}{{\sin A + \sin B}}$$$$ \Rightarrow \dfrac{{({R^2}4\sin A)\sin (\pi - (B + C)\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{({R^2}4{{\sin }^2}B)\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{({R^2}4{{\sin }^2}C)\sin (A - B)}}{{\sin A + \sin B}}
As we know sine in second quadrant is positive so we can write \sin (\pi - (B + C)) = \sin (B + C)$$$ \Rightarrow \dfrac{{({R^2}4\sin A)\sin (B + C)\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{({R^2}4{{\sin }^2}B)\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{({R^2}4{{\sin }^2}C)\sin (A - B)}}{{\sin A + \sin B}}$$ Now we will apply identity \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y $$ \Rightarrow \dfrac{{({R^2}4\sin A)({{\sin }^2}B - {{\sin }^2}C)}}{{\sin B + \sin C}} + \dfrac{{({R^2}4{{\sin }^2}B)\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{({R^2}4{{\sin }^2}C)\sin (A - B)}}{{\sin A + \sin B}}$$ Now we will apply algebraic identity i.e.{a^2} - {b^2} = (a - b)(a + b)$$$ \Rightarrow \dfrac{{({R^2}4\sin A)(\sin B - \sin C)(\sin B + \sin C)}}{{\sin B + \sin C}} + \dfrac{{({R^2}4{{\sin }^2}B)\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{({R^2}4{{\sin }^2}C)\sin (A - B)}}{{\sin A + \sin B}}$Now cancelling alike term from numerator and denominator we will get$ \Rightarrow ({R^2}4\sin A)(\sin B - \sin C) + \dfrac{{({R^2}4{{\sin }^2}B)\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{({R^2}4{{\sin }^2}C)\sin (A - B)}}{{\sin A + \sin B}}$Similarly other two parts of the equation can also be simplified by using these identities we have used above which will result in three similar forms of terms.$ \Rightarrow ({R^2}4\sin A)(\sin B - \sin C) + ({R^2}4\sin B)(\sin C - \sin A) + ({R^2}4\sin C)(\sin A - \sin B)$Now we will take${R^2}4$common from whole equation.$ \Rightarrow {R^2}4[(\sin A)(\sin B - \sin C) + (\sin B)(\sin C - \sin A) + (\sin C)(\sin A - \sin B)]$Now we will multiply the bracket terms so that we can further solve and simplify the equation.$ \Rightarrow {R^2}4[(\sin A\sin B - \sin A\sin C) + (\sin B\sin C - \sin B\sin A) + (\sin C\sin A - \sin C\sin B)]$Now we will cancel alike and opposite in sign terms from the above equation.$ \Rightarrow {R^2}4[0]$$ \Rightarrow 0$$
Hence it is proved that $\dfrac{{{a^2}\sin (B - C)}}{{\sin B + \sin C}} + \dfrac{{{b^2}\sin (C - A)}}{{\sin C + \sin A}} + \dfrac{{{c^2}\sin (A - B)}}{{\sin A + \sin B}} = 0$

Note: Students may likely make mistakes while applying so many identities at the same time so they must remember all important identities very cautiously. Also relating algebraic identities with trigonometric functions can sometimes become a little confusing but students should not get confused. The only thing which should be remembered is only taking care of degrees.