Question

If cot⁻¹1/x+cot⁻¹1/y+cot⁻¹1/z = π/2, then prove that xy+yz+zx=1

Answer 1

Proof :

$\mathrm{Given,\:cot^{-1}\frac{1}{x}+cot^{-1}\frac{1}{y}+cot^{-1}\frac{1}{z}=\frac{\pi}{2}}$

$\implies \mathrm{cot^{-1}\frac{\frac{1}{x}\frac{1}{y}-1}{\frac{1}{x}+\frac{1}{y}}+cot^{-1}\frac{1}{z}=\frac{\pi}{2}}$

$\implies \mathrm{cot^{-1}\frac{1-xy}{x+y}+cot^{-1}\frac{1}{z}=\frac{\pi}{2}}$

$\implies \mathrm{cot^{-1}\frac{\frac{1-xy}{x+y}\frac{1}{z}-1}{x+y+\frac{1}{z}}=\frac{\pi}{2}}$

$\implies \mathrm{\frac{\frac{1-xy-zx-yz}{(x+y)z}}{x+y+\frac{1}{z}}=cot\frac{\pi}{2}=0}$

$\implies \mathrm{1-xy-zx-yz=0}$

$\therefore \boxed{\mathrm{xy+yz+zx=1}}$

Hence, proved.

Answer 2

Step-by-step explanation:

$\huge\underline\mathfrak\purple{Solution}$