Question

In any $\Delta ABC,$ prove that:
$\dfrac{\left( {{b}^{2}}-{{c}^{2}} \right)}{{{a}^{2}}}\sin 2A+\dfrac{\left( {{c}^{2}}-{{a}^{2}} \right)}{{{b}^{2}}}\sin 2B+\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{{{c}^{2}}}\sin 2C=0$.

Answer 1

Hint: Use the formula of properties of triangle given by, $a=2R\sin A,b=2R\sin B\text{ and }c\text{=}2R\sin C$, where a, b and c are the lengths of the sides of the triangle opposite to vertex A, B and C respectively and ‘$R$’ is the circumradius of the circumcircle of the triangle. A, B and C also represent the angles of the triangle. Cancel the common terms and simplify the expression to prove the result.

Complete step-by-step answer:

We have been provided with the expression,
$\dfrac{\left( {{b}^{2}}-{{c}^{2}} \right)}{{{a}^{2}}}\sin 2A+\dfrac{\left( {{c}^{2}}-{{a}^{2}} \right)}{{{b}^{2}}}\sin 2B+\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{{{c}^{2}}}\sin 2C$
We need to prove that the value of this expression is zero.
Therefore,
$LHS=\dfrac{\left( {{b}^{2}}-{{c}^{2}} \right)}{{{a}^{2}}}\sin 2A+\dfrac{\left( {{c}^{2}}-{{a}^{2}} \right)}{{{b}^{2}}}\sin 2B+\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)}{{{c}^{2}}}\sin 2C$
Applying the formula of relation between side and the angle of a triangle, we get,
$a=2R\sin A,b=2R\sin B\text{ and }c=2R\sin C$. Therefore,
$\begin{align}
  & LHS=\dfrac{{{(2R\sin B)}^{2}}-{{(2R\sin C)}^{2}}}{{{(2R\sin A)}^{2}}}\times \sin 2A+\dfrac{{{(2R\sin C)}^{2}}-{{(2R\sin A)}^{2}}}{{{(2R\sin B)}^{2}}}\times \sin 2B \\
 & \text{ }+\dfrac{{{(2R\sin A)}^{2}}-{{(2R\sin B)}^{2}}}{{{(2R\sin C)}^{2}}}\times \sin 2C \\
 & LHS=\dfrac{4{{R}^{2}}{{\sin }^{2}}B-4{{R}^{2}}{{\sin }^{2}}C}{4{{R}^{2}}{{\sin }^{2}}A}\times \sin 2A+\dfrac{4{{R}^{2}}{{\sin }^{2}}C-4{{R}^{2}}{{\sin }^{2}}A}{4{{R}^{2}}{{\sin }^{2}}B}\times \sin 2B \\
 & \text{ }+\dfrac{4{{R}^{2}}{{\sin }^{2}}A-4{{R}^{2}}{{\sin }^{2}}B}{4{{R}^{2}}{{\sin }^{2}}C}\times \sin 2C \\
\end{align}$
Taking $4{{R}^{2}}$ common and cancelling from each term, we get,
$LHS=\dfrac{{{\sin }^{2}}B-{{\sin }^{2}}C}{{{\sin }^{2}}A}\times \sin 2A+\dfrac{{{\sin }^{2}}C-{{\sin }^{2}}A}{{{\sin }^{2}}B}\times \sin 2B+\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}B}{{{\sin }^{2}}C}\times \sin 2C$
Using the algebraic identity: ${{a}^{2}}-{{b}^{2}}=(a+b)\times (a-b)$, we get,
$\begin{align}
  & L.H.S=\dfrac{(\sin B-\sin C)(\sin B+\sin C)}{{{\sin }^{2}}A}\times \sin 2A+\dfrac{(\sin C-\sin A)(\sin C+\sin A)}{{{\sin }^{2}}B}\times \sin 2B \\
 & \text{ }+\dfrac{(\sin A-\sin B)(\sin A+\sin B)}{{{\sin }^{2}}C}\times \sin 2C \\
\end{align}$
Now, applying the formulas,
$\begin{align}
  & \sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right), \\
 & \sin a-\sin b=2\sin \left( \dfrac{a-b}{2} \right)\cos \left( \dfrac{a+b}{2} \right) \\
\end{align}$
We have,
$\begin{align}
  & L.H.S=\dfrac{2\sin \left( \dfrac{B-C}{2} \right)\cos \left( \dfrac{B+C}{2} \right)\times 2\sin \left( \dfrac{B+C}{2} \right)\cos \left( \dfrac{B-C}{2} \right)}{{{\sin }^{2}}A}\times \sin 2A \\
 & \text{ }+\dfrac{2\sin \left( \dfrac{C-A}{2} \right)\cos \left( \dfrac{C+A}{2} \right)\times 2\sin \left( \dfrac{C+A}{2} \right)\cos \left( \dfrac{C-A}{2} \right)}{{{\sin }^{2}}B}\times \sin 2B \\
 & \text{ }+\dfrac{2\sin \left( \dfrac{A-C}{2} \right)\cos \left( \dfrac{A+C}{2} \right)\times 2\sin \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)}{{{\sin }^{2}}C}\times \sin 2C \\
\end{align}$
Using the formula, $2\sin a\cos a=\sin 2a$, we get,
$L.H.S=\dfrac{\sin (B+C)\sin (B-C)}{{{\sin }^{2}}A}\times \sin 2A+\dfrac{\sin (C-A)\sin (C+A)}{{{\sin }^{2}}B}\times \sin 2B+\dfrac{\sin (A+B)\sin (A-B)}{{{\sin }^{2}}C}\times \sin 2C$
Now, we know that, in a $\Delta ABC,$$\angle A+\angle B+\angle C={{180}^{\circ }}$, therefore,
$\begin{align}
  & B+C={{180}^{\circ }}-A \\
 & A+C={{180}^{\circ }}-B \\
 & A+B={{180}^{\circ }}-C \\
\end{align}$
Using the above results, we get,
\[\begin{align}
  & LHS=\dfrac{\sin ({{180}^{\circ }}-A)\sin (B-C)}{{{\sin }^{2}}A}\times \sin 2A \\
 & \text{ }+\dfrac{\sin ({{180}^{\circ }}-B)\sin (C-A)}{{{\sin }^{2}}B}\times \sin 2B \\
 & \text{ }+\dfrac{\sin ({{180}^{\circ }}-C)\sin (A-B)}{{{\sin }^{2}}C}\times \sin 2C \\
\end{align}\]
Using the conversion: $\sin ({{180}^{\circ }}-\theta )=\sin \theta $, we have,
\[L.H.S=\dfrac{\sin A\sin (B-C)}{{{\sin }^{2}}A}\times \sin 2A+\dfrac{\sin B\sin (C-A)}{{{\sin }^{2}}B}\times \sin 2B+\dfrac{\sin C\sin (A-B)}{{{\sin }^{2}}C}\times \sin 2C\]
Again using, $\sin 2a=2\sin a\cos a$, we get,
\[\] \[\begin{align}
  & LHS=\dfrac{\sin A\sin (B-C)}{{{\sin }^{2}}A}\times 2\sin A\cos A+\dfrac{\sin B\sin (C-A)}{{{\sin }^{2}}B}\times 2\sin B\cos B+\dfrac{\sin C\sin (A-B)}{{{\sin }^{2}}C}\times 2\sin C\cos C \\
 & LHS=\dfrac{2{{\sin }^{2}}A\cos A\sin (B-C)}{{{\sin }^{2}}A}+\dfrac{2{{\sin }^{2}}B\cos B\sin (C-A)}{{{\sin }^{2}}B}+\dfrac{2{{\sin }^{2}}C\cos C\sin (A-B)}{{{\sin }^{2}}C} \\
\end{align}\]
Now, cancelling the common terms we get,
\[L.H.S=2\left( \cos A\sin (B-C)+\cos B\sin (C-A)+\cos C\sin (A-B) \right)\]
Using the expansion formula: $\sin (a-b)=\sin a\cos b-\cos a\sin b$, we get,
\[LHS=2\left( \begin{align}
  & \cos A\times \left( \sin B\cos C-\cos B\sin C \right) \\
 & +\cos B\times \left( \sin C\cos A-\cos C\sin A \right) \\
 & +\cos C\times \left( \sin A\cos B-\cos A\sin B \right) \\
\end{align} \right)\]
\[LHS=2\left( \begin{align}
  & \cos A\sin B\cos C \\
 & -\cos A\cos B\sin C \\
 & +\cos B\sin C\cos A \\
 & -\cos B\cos C\sin A \\
 & +\cos C\sin A\cos B \\
 & -\cos C\cos A\sin B \\
\end{align} \right)\]
\[LHS=2\left( \begin{align}
  & (\cos A\sin B\cos C-\cos A\sin B\cos C) \\
 & +(\cos A\cos B\sin C-\cos A\cos B\sin C) \\
 & +(\cos B\cos C\sin A-\cos B\cos C\sin A) \\
\end{align} \right)\]
LHS = 0 = RHS
Hence proved

Note: There can be an alternate method to prove the above result. We can assign any particular value to the angle A, B and C and then substitute the value of the sine of the assumed angle according to the expression and then simplify the expression. But this alternate method is recommended when we have been provided with different options and we have to find the value of the expression.