if cos-1x+cos-1y+cos​-1z=pi and x+y+z=3/2 prove that x=y=z

Please give me a proper method of solving this question, DONT assume x=y=z and then solve
Properly solve using the given conditions and NOT the other way round

Dear Student,
Please find below the solution to the asked query:


$$\begin{gathered} \mathrm{Now} \\ \mathrm{x}+\mathrm{y}+\mathrm{z}=\frac{3}{2} \\ \mathrm{Square} \mathrm{both} \mathrm{sides} \\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+2\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)=\frac{9}{4} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=\frac{9}{4}-2\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ \mathrm{Put} \mathrm{in} {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+2\mathrm{xyz}=1 \\ \Rightarrow \frac{9}{4}-2\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)+2\mathrm{xyz}=1 \\ \Rightarrow 9-8\mathrm{xy}-8\mathrm{yz}-8\mathrm{zx}+8\mathrm{xyz}=4 \\ \Rightarrow 5-8\mathrm{xy}-8\mathrm{yz}-8\mathrm{zx}+8\mathrm{xyz}=0 \\ \mathrm{Which} \mathrm{holds} \mathrm{good} \mathrm{for} \mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{1}{2} \\ \mathrm{i}.\mathrm{e}. \\ 5-8\times \frac{1}{2}\times \frac{1}{2}-8\times \frac{1}{2}\times \frac{1}{2}-8\times \frac{1}{2}\times \frac{1}{2}+8\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} \\ =5-2-2-2+1 \\ =0 \\ \mathrm{i}.\mathrm{e}. \mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{1}{2} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards