In a `DeltaABC, cos A+cos B+cosC` belongs to the interval

Options

A. `(1//2,3//2)`
B. `(1//,3//2)`
C. `(3//2,2)`
D. none of these

Answer

B

Explanation

We have,
`cosA+cos B+cosC`
`=2cos""(A+B)/(2)cos""(A-C)/(2)+1-2sin""C/2`
`=2sin""C/2{cos""(A-B)/(2)-sin""(C)/(2)}+1`
`le2sin""(C)/(2){1-sin""(C)/(2)}+1" "...(i)`
`" "[because"Max, value of "cos""(A-B)/(2)is 1]`
`=1-2{sin^(2)""(C)/(2)-sin""(C)/(2)}`
`=1-2{(sin ""(C)/(2)-1/2)^(2)-1/4}`
`=3/2-2(sin""(C)/(2)-1/2)^(2)`
Thus, we have
`cosA+cosB+cosC le3/2-2(sin""(C)/(2)-1/2)^(2)" "...(ii)`
`impliescosA+cosB+cosCle3/2`
It is evident from (i) and (ii) that
`cosA+cosB+cosC3/2,`
`if cos""(A-B)/(2)=1and sin""(C)/(2)-1/2=0`
`impliesA-B=0andC=60^(@)`
`impliesA=Band C=60^(@)`
`impliesA=B=C60^(@)`
`impliesDeltaABC` is equilateral.
`becausecosA+cosB+cosC=4sin""A/2sin""B/2sin""C/2+1`
`and, 4 sinA//2sinB//2sinC//2gt0`
`thereforecosA+cosB+cosCgt1`
Thus, `1ltcosA+cosB+cosCle3/2`
Hence, option (b) is correct.