If $(1 - s i n A) (1 - s i n B) (1 - s i n C) = (1 + s i n A) (1 + s i n B) (1 + s i n C) = k$

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Solution

$(1 - s i n A) (1 - s i n B) (1 - s i n C) = (1 + s i n A) (1 + s i n B) (1 + s i n C) = k$
this is possible only when
$(1 - s i n A) = (1 + s i n A)$
$(1 - s i n B) = (1 + s i n B)$
$(1 - s i n C) = (1 + s i n C)$
now,
$(1 - s i n A) = (1 + s i n A)$
$s i n A = 0$
$A = 0$
in the same way $B = C = 0$
now,
$k = (1 - 0) (1 - 0) (1 - 0) = (1 + 0) (1 + 0) (1 + 0) = 1 x 1 x 1 = 1$
hence,
$k = 1$

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