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Question

The value of tan1a(a+b+c)bc+tan1b(a+b+c)ca+tan1c(a+b+c)ab where a,b,c>0 is

A
π4
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B
π2
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C
π
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D
0
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Solution

The correct option is C π
Let S=tan1a(a+b+c)bc+tan1b(a+b+c)ca+tan1c(a+b+c)ab
Let x=a(a+b+c)bc, y=b(a+b+c)ca and z=c(a+b+c)ab

x+y+z=a+b+c(abc+bca+cab)
=a+b+c(a+b+cabc)=(a+b+c)3/2(abc)1/2
xyz=a(a+b+c)bcb(a+b+c)cac(a+b+c)ab=(a+b+c)3/2(abc)1/2
x+y+z=xyzS=π

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