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 `(d a)/(c a sA)+(d b)/(cosB)+(d c)/(cosC)=.`

Options

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

Answer

A

Explanation

We know that
` rArr " "a=2R sin A, b= 2R sin B and c= 2R sin C`
`rArr (da)/(dA)= 2R cos A,(db)/(dB) = 2R cos B and (dc)/(dC) = 2 R cos C`
But, ` da=(da)/(dA) dA, db=(db)/(dB)dB and dc=(dc)/(dC)dC`
`da=2R cos A" "dA, db= 2R cos B " "dB and dc=2R cos C" "dc`
`rArr (da)/cos A+(db)/cos B+(dc)/cos C= 2R (dA+dB=dC)`
`rArr" "(da)/cos A+(db)/cos B+(dc)/cos C= 2Rd (A+B+C)= 2R" "d(pi)`
`rArr" "(da)/cos A+(db)/cos B+(dc)/cosC = 2R" "(0)=0 " "[:' A+B+C=pi]`