If a,b,c be positive real number and the value of θ=tan-1a(a+b+c)bc+tan-1b(a+b+c)ca+tan-1c(a+b+c)ab, tanθ:
0
1
a+b+cabc
Noneofthese
Finding the value of
Given, a,b,c be positive real number
Let, x2=a+b+cabc
Now,
θ=tan-1a(a+b+c)bc+tan-1b(a+b+c)ca+tan-1c(a+b+c)abθ=tan-1a2x2+tan-1b2x2+tan-1c2x2θ=tan-1ax+tan-1bx+tan-1cxθ=tan-1[[ax+bx+cx–abcx3]1–x2(ab+bc+ca)]tanθ=x[(a+b+c)-abc(a+b+c)abc]1–x2(ab+bc+ca){x2=a+b+cabc}tanθ=0
Hence, correct option is (A)