If a triangle ABC, inscribed in a fixed circle, be slightly varied in such away as to have its vertices always on the circle, then show that `(d a)/(c a sA)+(d b)/(cosB)+(d c)/(cosC)=0.`

Explanation

We have to show that

`(d a)/(c a sA)+(d b)/(cosB)+(d c)/(cosC)=0.`
`LHS=(d a)/(c a sA)+(d b)/(cosB)+(d c)/(cosC)......(1)`
we know that,
`a=2RsinA , b=2R sinB , c=2RsinC `
so,
`(da)/(dA)=2Rcos A,(db)/(dB)=2Rcos B,(dc)/(dC)=2Rcos C`
`=>da=2RcosA dA ,db=2RcosB dB,dc=2RcosC dC`
substituting these value in `(1)` we get,
`(2RcosA dA)/(cos A)+(2RcosB dB)/(cos B)+(2RcosC dC)/(cos C)`
`=2R dA+2R dB +2RdC`
`=2R(dA+dB+dC)`
`=2R*d(A+B+C)`
`=2R*d pi`
`=0=RHS`