Relating variables .

Algebra Level 3

If $\frac { x-y }{ 1+xy } +\frac { y-z }{ 1+yz } +\frac { z-x }{ 1+zx } =0$ , then which of the following statements are true?

Note: Try to prove all the parts if you get the correct answer .

all variables are equal and equal to zero all variables are equal None of the above some two variables must be equal

1 solution

Anish Kelkar
Jun 29, 2014

Let us substitute $x=tanA\quad y=tanB\quad z=tanC$ . $\frac { x-y }{ 1+xy } +\frac { y-z }{ 1+yz } +\frac { z-x }{ 1+zx } =tan(A-B)+tan(B-C)+tan(C-A)=0\quad \quad \\ \\ \Longrightarrow \quad \frac { sin(A-C) }{ cos(A-B)cos(B-C) } +\frac { sin(C-A) }{ cos(C-A) } =0\\ \\ For\quad sin(A-C)\neq 0\quad \\ \quad \quad \quad \quad \quad \quad cos(C-A)\quad =\quad cos(A-C)\quad +cos(A+C-2B)\quad \\ \Rightarrow \quad \quad \quad cos(C-A)=\quad cos(A+C-2B)\quad \\ So,\quad \\ \quad \quad \quad either\quad \quad \quad C-A\quad =\quad A+C-2B\quad \quad \Rightarrow \quad A=B\\ \quad \quad \quad \quad \quad or\quad \quad \quad \quad C-A\quad =\quad -(A+C-2B)\quad \Rightarrow \quad C=B\\ \quad \\ For\quad sin(A-C)=0\quad \Rightarrow \quad A=C\quad \quad \\ Hence\quad Prooved\quad .$

How did you get $\cos (C-A) = \cos (A-C) + \cos (A+C - 2B)$ ?

I believe that your option should be "Some 2 of the variables are equal", as opposed to "Any 2 of the variables are equal". The latter implies that $a=b$ AND $b=c$ AND $c =a$ . I have updated it accordingly.

Calvin Lin Staff - 6 years, 11 months ago

thanks for the correction @Calvin Lin

Anish Kelkar - 6 years, 10 months ago

Relating variables .

1 solution

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