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(A) Let $s^2 = \frac{a+b+c}{abc}$. Then the expression becomes $	heta = 	an^{-1}\sqrt{a^2 s^2} + 	an^{-1}\sqrt{b^2 s^2} + 	an^{-1}\sqrt{c^2 s^2}$.

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(A) Let $s^2 = \frac{a+b+c}{abc}$. Then the expression becomes $	heta = 	an^{-1}\sqrt{a^2 s^2} + 	an^{-1}\sqrt{b^2 s^2} + 	an^{-1}\sqrt{c^2 s^2}$.Since $a, b, c$ are positive,$	heta = 	an^{-1}(as) + 	an^{-1}(bs) + 	an^{-1}(cs)$.Using the formula $	an^{-1} x + 	an^{-1} y + 	an^{-1} z = 	an^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}ight)$:$	an 	heta = \frac{as+bs+cs - (as)(bs)(cs)}{1 - (asbs + bscs + csas)} = \frac{s(a+b+c) - s^3(abc)}{1 - s^2(ab+bc+ca)}$.Substitute $s^2 = \frac{a+b+c}{abc}$,which implies $s^2(abc) = a+b+c$:$	an 	heta = \frac{s(a+b+c) - s(s^2 abc)}{1 - s^2(ab+bc+ca)} = \frac{s(a+b+c) - s(a+b+c)}{1 - s^2(ab+bc+ca)} = \frac{0}{1 - s^2(ab+bc+ca)} = 0$.Alternatively,let $a=b=c=1$. Then $	heta = 	an^{-1}\sqrt{3} + 	an^{-1}\sqrt{3} + 	an^{-1}\sqrt{3} = 3 	imes 60^\circ = 180^\circ = \pi$. Thus,$	an 	heta = 	an \pi = 0$.

If $a, b, c$ are positive real numbers and the value of $\theta = \tan^{-1}\sqrt{\frac{a(a+b+c)}{bc}} + \tan^{-1}\sqrt{\frac{b(a+b+c)}{ca}} + \tan^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$,then $\tan \theta$ is equal to

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