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Question

# if a, b, c be positive real numbers and the value of θ=tan−1√a(a+b+c)bc+tan−1√b(a+b+c)ca+tan−1√c(a+b+c)ab then tanθ is equal to:

A
0
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B
1
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C
a+b+cabc
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D
None of these
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Solution

## The correct option is A 0Let k=√a+b+cabc⟹k2=a+b+cabc⟹abck2=a+b+cθ=tan−1√a(a+b+c)bc+tan−1√b(a+b+c)ca+tan−1√c(a+b+c)ab =tan−1√a2(a+b+c)abc+tan−1√b2(a+b+c)abc+tan−1√c2(a+b+c)abc =tan−1ak+tan−1bk+tan−1ck =tan−1(k(a+b+c)−abck31−(ab+bc+ca)k2) =tan−1(k((a+b+c)−abck2)1−(ab+bc+ca)k2) =tan−1(k((a+b+c)−(a+b+c))1−(ab+bc+ca)k2) =tan−1(0)⟹tanθ=0

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