If $x + y + z = x y z$ , then prove that.
For any three angles A, B, C, prove that $tan (B - C) + tan (C - A) + tan (A - B) = tan (B - C) tan (C - A) tan (A - B)$ .

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Solution

Put $B - C = p, C - A = q, A - B = r$
$∴ p + q + r = 0$
or $tan (p + q + r) = tan 0^{o} = 0$
or $\frac{S_{1} - S_{3}}{1 - S_{2}} = 0$
$∴ S_{1} = S_{3}$
or $tan p + tan q + tan r = tan p tan q tan r$
Now put the values of p, q and r.
$∴ tan (B - C) + tan (C - A) + tan (A - B) = tan (B - C) tan (C - A) tan (A - B)$ .

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