Question

If \[\tan^{-1}\left(\frac{x}{2}\right)+\tan^{-1}\left(\frac{y}{2}\right)+\tan^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2}\]  then xy + yz + zx =

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Answer 1

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Explanation:

\[\tan^{-1}\left(\frac{x}{2}\right)+\tan^{-1}\left(\frac{y}{2}\right)+\tan^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2}\] 

Let \[\frac{x}{2}=\mathrm{a,\frac{y}{2}=b,\frac{z}{2}=c}\]

\[\tan^{-1}\mathrm{a}+\tan^{-1}\mathrm{b}+\tan^{-1}\mathrm{c}=\frac{\pi}{2}\]

\[\tan^{-1}\left(\frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}\right)+\tan^{-1}\mathrm{c}=\] \[\frac{\pi}{2}\]

\[\therefore\quad\tan^{-1} \begin{bmatrix} \frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{a}\mathrm{b}}+\mathrm{c} \\ -\frac{\left(\mathrm{a}+\mathrm{b}\right)\mathrm{c}}{1-\mathrm{a}\mathrm{b}} \end{bmatrix}=\frac{\pi}{2}\]

\[\begin{array} {ccc} & & \mathrm{} \\ \therefore & & \frac{\mathrm{a+b+c}\left(1-\mathrm{ab}\right)}{1-\mathrm{ab-ac-bc}}=\tan\frac{\pi}{2} \end{array}\]

\[\therefore\quad\frac{\mathrm{a+b+c\left(1-ab\right)}}{1-\mathrm{ab-ac-bc}}=\mathrm{tan}\frac{\pi}{2}\]

\[\therefore\quad1-\mathrm{ab}-\mathrm{ac}-\mathrm{bc}=0\quad\ldots\left\lfloor\because\mathrm{tan}\frac{\pi}{2}=\infty\right\rfloor\]

\[\therefore\quad\mathrm{ab+bc+ac=1}\]

\[\therefore\quad\frac{x}{2}\left(\frac{y}{2}\right)+\frac{y}{2}\left(\frac{z}{2}\right)+\frac{x}{2}\left(\frac{z}{2}\right)=1\]

\[\therefore\quad\frac{x}{2}\left(\frac{y}{2}\right)+\frac{y}{2}\left(\frac{z}{2}\right)+\frac{x}{2}\left(\frac{z}{2}\right)=1\]

\(\therefore\quad xy+yz+xz=4\)