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) \[\pm \cos A\sin B\sin C\]
Correct Answer: B
Solution :
Multiplying both sides by \[(1-\sin A)(1-\sin B)(1-\sin C)\], we have, \[(1-{{\sin }^{2}}A)(1-{{\sin }^{2}}B)(1-{{\sin }^{2}}C)\] \[={{(1-\sin A)}^{2}}{{(1-\sin B)}^{2}}{{(1-\sin C)}^{2}}\] Þ \[(1-\sin A)(1-\sin B)(1-\sin C)=\pm \cos A\cos B\cos C\] Similarly, \[(1+\sin A)(1+\sin B)(1+\sin C)=\pm \cos A\cos B\cos C\].
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