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3.Trigonometrical Ratios, Functions and Identities
easy
If $(1 + \sin A)(1 + \sin B)(1 + \sin C)$$ = (1 - \sin A)(1 - \sin B)(1 - \sin C),$ then each side is equal to
A
$ \pm \sin A\sin B\sin C$
B
$ \pm \cos A\cos B\cos C$
C
$ \pm \sin A\cos B\cos C$
D
$ \pm \cos A\sin B\sin C$
Solution
(b) Multiplying both sides by $(1 – \sin A)(1 – \sin B)(1 – \sin C)$,
we have, $(1 – {\sin ^2}A)(1 – {\sin ^2}B)(1 – {\sin ^2}C)$
$ = {(1 – \sin A)^2}{(1 – \sin B)^2}{(1 – \sin C)^2}$
==> $(1 – \sin A)(1 – \sin B)(1 – \sin C) = \pm \cos A\cos B\cos C$
Similarly, $(1 + \sin A)(1 + \sin B)(1 + \sin C) = \pm \cos A\cos B\cos C$.
Standard 11
Mathematics
