Question

In any triangle abc prove that (b^2 -c^2)/a^2 sin2a + (c^2 -a^2)/b^2 sin 2b + (a^2 -b^2)/c^2 sin 2 =0

Answer 1

We know that in a triangle $a=k\sin A,b=k\sin B,c=k\sin C$.

So

$\frac{\sin 2A}{a^2} =\frac{2}{a^2}  \sin A \cos A=\frac{2a}{ka^2} \frac{b^2+c^2-a^2}{2bc} =\frac{b^2+c^2-a^2}{2kabc}$.

Similarly,

[tex]\frac{\sin 2B}{b^2} =\frac{c^2+a^2-b^2}{2kabc} \\ \frac{\sin 2C}{c^2} =\frac{a^2+b^2-c^2}{2kabc}[/tex].

Thus,

$LHS=(b^2 -c^2)\frac{\sin 2A}{a^2} +(c^2 -b^2)\frac{\sin 2B}{b^2}+(a^2 -c^2)\frac{\sin 2C}{c^2}$

[tex]LHS=(b^2 -c^2)\frac{b^2+c^2-a^2}{2kabc} +(c^2 -b^2)\frac{c^2+a^2-b^2}{2kabc}+(a^2 -b^2)\frac{a^2+b^2-c^2}{2kabc}\\ LHS=\frac{b^4-c^4-a^2(b^2 -c^2)}{2kabc} +\frac{c^4-a^4-b^2(c^2 -b^2)}{2kabc}+\frac{a^4-b^4-c^2(a^2 -b^2)}{2kabc}\\ LHS=\frac{b^4-c^4-a^2(b^2 -c^2)+c^4-a^4-b^2(c^2 -b^2)+a^4-b^4-c^2(a^2 -b^2)}{2kabc} \\ LHS=\frac{0}{2kabc} =0=RHS[/tex].

The proof is complete.

Answer 2

Answer:

Step-by-step explanation:

we know sine rule

$\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}=k$

we get

a= k sinA , b=ksinB , c=ksinC

the value of $\frac{\sin 2A}{a^2} =\frac{2}{a^2} \sin A \cos A=\frac{2a}{ka^2} \frac{b^2+c^2-a^2}{2bc} =\frac{b^2+c^2-a^2}{2kabc}.$

similarly

$\frac{\sin 2B}{b^2} =\frac{c^2+a^2-b^2}{2kabc}\ \\\\frac{\sin 2C}{c^2} =\frac{a^2+b^2-c^2}{2kabc}.$

LHS=$(b^2 -c^2)\frac{\sin 2A}{a^2} +(c^2 -b^2)\frac{\sin 2B}{b^2}+(a^2 -c^2)\frac{\sin 2C}{c^2}$

$=(b^2 -c^2)\frac{b^2+c^2-a^2}{2kabc} +(c^2 -b^2)\frac{c^2+a^2-b^2}{2kabc}+(a^2 -b^2)\frac{a^2+b^2-c^2}{2kabc}\\\\</p><p>=\frac{b^4-c^4-a^2(b^2 -c^2)}{2kabc} +\frac{c^4-a^4-b^2(c^2 -b^2)}{2kabc}+\frac{a^4-b^4-c^2(a^2 -b^2)}{2kabc}\\\\</p><p>=\frac{b^4-c^4-a^2(b^2 -c^2)+c^4-a^4-b^2(c^2 -b^2)+a^4-b^4-c^2(a^2 -b^2)}{2kabc} \\\\</p><p>=\frac{0}{2kabc} =0$