1.

For any triangle ABC, prove that`(b^2-c^2)/(a^2)sin2A+(c^2-a^2)/(b^2)sin2B+(a^2-b^2)/(c^2)sin2C=0`

Answer» For any triangle ABC, we have,`sinA/a = sinB/b = sinC/c = k`->(1)
`cos A = (b^2+c^2-a^2)/(2bc), cosB = (c^2+a^2-b^2)/(2ca), cos C = (a^2+b^2-c^2)/(2ab)`->(2)
Now, we will solve our equation.
`L.H.S = (b^2-c^2)/a^2 *2sinAcosA+(c^2-a^2)/b^2*2sinBcosB+(a^2-b^2)/c^2*2sinCcosC`
From (1),
`=(b^2-c^2)/a^2*2kacosA+(c^2-a^2)/b^2*2kbcosB+(a^2-b^2)/c^2*2kccosC`
`=2k((b^2-c^2)/a*(b^2+c^2-a^2)/(2bc)+(c^2-a^2)/b*(c^2+a^2-b^2)/(2ca)+(a^2-b^2)/c*(a^2+b^2-c^2)/(2ab))`
`=k/(abc)(b^4-c^4-a^2(b^2-c^2)+c^4-a^4-b^2(c^2-a^2)+a^4-b^4-c^2(a^2-b^2))`
`=k/(abc)(0) = 0= R.H.S.`


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