If $ an ^{-1} x+ an ^{-1} y+ an ^{-1} z=\frac{\pi}{2}, \quad x, y, z0, x y

Question

If $	an ^{-1} x+	an ^{-1} y+	an ^{-1} z=\frac{\pi}{2}, \quad x, y, z>0, x y < 1, \quad$ then the value of $x y+y z+z x=$

Aicharya · Accepted Answer

$	an ^{-1} x+	an ^{-1} y+	an ^{-1} z=\frac{\pi}{2}$ $	herefore\left(	an ^{-1} x+	an ^{-1} yight)=\left(\frac{\pi}{2}-	an ^{-1} zight)$ $	herefore 	an ^{-1}\left(\frac{x+y}{1-x y}ight)=\cot ^{-1} z=	an ^{-1}\left(\frac{1}{z}ight)$ $	herefore \frac{x+y}{1-x y}=\frac{1}{z} \Rightarrow x z+y z=1-x y$ $	herefore x y+y z+z x=1$

If $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}, \quad x, y, z>0, x y < 1, \quad$ then the value of…

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