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

Options

A. 0
B. 1
C. -1
D. none of these

Answer

A

Explanation

We are given that the side c angle C remain constant.
`:." "C/(sir" "C)=k(constant)` ` rArr a/(sir" "A)=b/(sin B)=k " "[:' a/(sin A)=b/(sin B)=c/(sin C)] `
`rArr " "a=k sin A and b=k cos B`
`rArr " "(da)/(dA)=k cos A and (db)/(dB)=k cos B`
Now, `da=(da)/(dA).dA rArrda=k cos A.dA rArr (da)/(cos A)= k" "dA`
and, `db=(db)/(dB).dB rArr=k cos B. dB rArr (db)/(cos B) =k" "dB'
`:." "(da)/(cos A)+(db)/(cos B)= k dA + K dB = k dB = kd(A+B) = k d (pi-C)`
`(da)/(cos A)+ (db)/(cos B) = k (0)=0" "{:[(.:' pi-C ="constant"), (.:.d(pi-C)=0)]:}`