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Miscellaneous Exercise 3 · Q137

Q.If tan⁡−1x+tan⁡−1y+tan⁡−1z=π2\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \dfrac{\pi}{2} then show that xy+yz+zx=1xy + yz + zx = 1.

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tan⁡−1x+tan⁡−1y=π2−tan⁡−1z\tan^{-1}x+\tan^{-1}y=\dfrac{\pi}{2}-\tan^{-1}z. Taking tangent of both sides: tan⁡(tan⁡−1x+tan⁡−1y)=tan⁡(π2−tan⁡−1z)=cot⁡(tan⁡−1z)=1z\tan(\tan^{-1}x+\tan^{-1}y)=\tan\left(\dfrac{\pi}{2}-\tan^{-1}z\right)=\cot(\tan^{-1}z)=\dfrac1z. L.H.S.: $\dfrac{x+y}{1-xy}=\dfrac1z\ …

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