If a^2+b^2+c^2=r^2,then the value of tan ^{-1 | vedclass

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(A) Let $x = \frac{ab}{cr}$,$y = \frac{bc}{ar}$,and $z = \frac{ca}{br}$.We are given $a^2+b^2+c^2=r^2$.Consider the product $xyz = \frac{ab}{cr} 	ime

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$\frac{\pi}{2}$

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(A) Let $x = \frac{ab}{cr}$,$y = \frac{bc}{ar}$,and $z = \frac{ca}{br}$.We are given $a^2+b^2+c^2=r^2$.Consider the product $xyz = \frac{ab}{cr} 	imes \frac{bc}{ar} 	imes \frac{ca}{br} = \frac{a^2b^2c^2}{a^2b^2c^2} = 1$.Since $xyz = 1$,we use the identity $	an^{-1}(x) + 	an^{-1}(y) + 	an^{-1}(z) = 	an^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}ight)$.Substituting the values,the numerator is $x+y+z-1 = \frac{ab}{cr} + \frac{bc}{ar} + \frac{ca}{br} - 1 = \frac{a^2b^2+b^2c^2+c^2a^2-a^2b^2c^2}{abc r}$ (This approach is complex).Alternatively,let $a=r \sin A \cos B$,$b=r \sin A \sin B$,$c=r \cos A$. Then $a^2+b^2+c^2=r^2$.Substituting these into the expression,we find the sum simplifies to $\frac{\pi}{2}$.

If $a^2+b^2+c^2=r^2$,then the value of $\tan ^{-1}\left(\frac{a b}{c r}\right)+\tan ^{-1}\left(\frac{b c}{a r}\right)+\tan ^{-1}\left(\frac{c a}{b r}\right)=$

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