If in a triangle `A B C ,(1+cosA)/a+(1+cosB)/b+(1_(cosC))/c` `=(k^2(1+cosA)(1+cosB)(1+cosC)/(a b c)` , then `k` is equal to `1/(2sqrt(2)R)` (b) `2R` (c) `1/R` (d) none of these

Explanation

`(1+cosA)/a+(1+cosB)/b +(1+cosC)/c = k^2((1+cosA)/a*(1+cosB)/b*(1+cosC)/c)`

`=>(2cos^2(A/2))/(2RsinA)+(2cos^2(B/2))/(2RsinB)+(2cos^2(C/2))/(2RsinC) = k^2((2cos^2(A/2))/(2RsinA)*(2cos^2(B/2))/(2RsinB)*(2cos^2(C/2))/(2RsinC))`

`=>(cos^2(A/2))/(2Rsin(A/2)cos(A/2))+(cos^2(B/2))/(2Rsin(B/2)cos(B/2))+(cos^2(C/2))/(2Rsin(C/2)cos(C/2)) = k^2((cos^2(A/2))/(2Rsin(A/2)cos(A/2))*(cos^2(B/2))/(2Rsin(B/2)cos(B/2))*(cos^2(C/2))/(2Rsin(C/2)cos(C/2)))`

`=>1/(2R)[cot(A/2)+cot(B/2)+cot(C/2)] = k^2/(8R^3)[cot(A/2)cot(B/2)cot(C/2)]`

By comparing coefficients,

`=>1/(2R) = k^2/(8R^3)`

`=>k^2 = 4R^2`

`=>k = 2R`

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