If in a triangle `A B C` , the side `c` and the angle `C` remain constant, while the remaining elements are changed slightly, using differentials show that `(d a)/(cosA)+(d b)/(cosB)=0`

Explanation

We know that
`a/(sin A) = b/(sin B) = c/(sin C)`
But a,b and C are constant.
Therefore `a/(sin A)= b/(sin B) = k`
` rArr a=k sin A and b=k sin B`
` rArr (da)/(dA) = k cos A and (db)/(dB) k cos B`
`rArr (da)/(cos A) = k * dA and (db)/(cos B) = k * dB`
`rArr (da)/(cos A) + (db)/(cos B) = k(dA+ dB)`
` = k d (A+B) - k d (pi - C)`
` = k * (0)`
`=0`
` (da)/(cos A) + (db)/(cos B) = 0`.