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Q): If $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=2$ then find the value of $(x+y+z)(x-y-z)$.

Ans): First of all let me tell everyone that this question is being asked today by the professors of Mathematics of my University. The way by which they solved this question is that they took into consideration $\sin^{-1}x=\sin^{-1}y=\sin^{-1}z=\frac{2}{3}$. Therefore from this step we are getting $x=y=z=\sin(\frac{2}{3})$. Now they used these values of $x,y,z$ to find the value of $(x+y+z)(x-y-z)$. I have understood the method and it is a good method.

Doubt: But here in this question I can't understand that why I have to always consider $x=y=z=\sin(\frac{2}{3})$, for this question. Here clearly many cases can arise. For example: if $\sin^{-1}x=1$ and $\sin^{-1}y=\sin^{-1}z=\frac{1}{2}$, then also $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=2$. Again let's say for example: $\sin^{-1}x=\frac{1}{3}$, $\sin^{-1}y=\frac{2}{3}$ and $\sin^{-1}z=1$, then also we can say that $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=2$. Similarly in this way many cases will arise. If the question was : "find the value of $(x+y+z)(x-y-z)$ if $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=3$", then one and only one case is possible and that is $\sin^{-1}x=\sin^{-1}y=\sin^{-1}z=1$. But for this question many cases are possible. So, please help me out with this question.

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    $\begingroup$ Yes,There need to be extra conditions to solve for exact values,Otherwise,Many such cases are possible $\endgroup$ Nov 29, 2023 at 11:56
  • $\begingroup$ What's the source of the problem, the right hand side is not $$\pi$$ by any chance? $\endgroup$ Dec 12, 2023 at 15:12

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