If a,b,c are positive then tan−1a(a+b+c)bc+tan−1b(a+b+c)ca+tan−1c(a+b+c)ab=

tan−1a(a+b+c)bc+tan−1b(a+b+c)ca+tan−1c(a+b+c)ab Let (a+b+c)=t =tan−1atbc+tan−1btac+tan−1ctab =tan−1(atbc+btac+ctab−atbc.btac.ctab1−(atbc.btac+btac.ctab+ctab.atbc)) =tan−1[t(a+b+c)abc−ttabc1−(tc+ta+tb)] =tan−1[ttabc−ttabc1−(tc+ta+tb)]=tan−1(0)=π

If a,b,c are positive then tan−1a(a+b+c)bc+tan−1b(a+b+c)ca+tan−1c(a+b+c)ab=

tan−1a(a+b+c)bc+tan−1b(a+b+c)ca+tan−1c(a+b+c)ab Let (a+b+c)=t =tan−1atbc+tan−1btac+tan−1ctab =tan−1(atbc+btac+ctab−atbc.btac.ctab1−(atbc.btac+btac.ctab+ctab.atbc)) =tan−1[t(a+b+c)abc−ttabc1−(tc+ta+tb)] =tan−1[ttabc−ttabc1−(tc+ta+tb)]=tan−1(0)=π

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If $a, b, c$ are positive then $\tan^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + \tan^{- 1} \sqrt{\frac{b (a + b + c)}{c a}} + \tan^{- 1} \sqrt{\frac{c (a + b + c)}{a b}} =$

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π

\frac{3 π}{2}

\frac{3 π}{4}

answer is A.

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Detailed Solution

\tan^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + \tan^{- 1} \sqrt{\frac{b (a + b + c)}{c a}} + \tan^{- 1} \sqrt{\frac{c (a + b + c)}{a b}}

$Let (a + b + c) = t$

$= \tan^{- 1} \sqrt{\frac{a t}{b c}} + \tan^{- 1} \sqrt{\frac{b t}{a c}} + \tan^{- 1} \sqrt{\frac{c t}{a b}}$

$= \tan^{- 1} (\frac{\sqrt{\frac{a t}{b c}} + \sqrt{\frac{b t}{a c}} + \sqrt{\frac{c t}{a b}} - \sqrt{\frac{a t}{b c}} . \sqrt{\frac{b t}{a c}} . \sqrt{\frac{c t}{a b}}}{1 - (\sqrt{\frac{a t}{b c}} . \sqrt{\frac{b t}{a c}} + \sqrt{\frac{b t}{a c}} . \sqrt{\frac{c t}{a b}} + \sqrt{\frac{c t}{a b}} . \sqrt{\frac{a t}{b c}})})$

$= \tan^{- 1} [\frac{\frac{\sqrt{t} (a + b + c)}{\sqrt{a b c}} - \frac{t \sqrt{t}}{\sqrt{a b c}}}{1 - (\frac{t}{c} + \frac{t}{a} + \frac{t}{b})}]$

$= \tan^{- 1} [\frac{\frac{t \sqrt{t}}{\sqrt{a b c}} - \frac{t \sqrt{t}}{\sqrt{a b c}}}{1 - (\frac{t}{c} + \frac{t}{a} + \frac{t}{b})}] = \tan^{- 1} (0) = π$