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Question: If \((1 + \sin A)(1 + \sin B)(1 + \sin C) = (1 - \sin A)(1 - \sin B)(1 - \sin C),\) then each side i...

If (1+sin⁡A)(1+sin⁡B)(1+sin⁡C)=(1−sin⁡A)(1−sin⁡B)(1−sin⁡C),(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

±sin⁡Asin⁡Bsin⁡C\pm \sin A\sin B\sin C

B

±cos⁡Acos⁡Bcos⁡C\pm \cos A\cos B\cos C

C

±sin⁡Acos⁡Bcos⁡C\pm \sin A\cos B\cos C

D

±cos⁡Asin⁡Bsin⁡C\pm \cos A\sin B\sin C

Answer

±cos⁡Acos⁡Bcos⁡C\pm \cos A\cos B\cos C

Explanation

Solution

Multiplying both sides by

(1−sin⁡A)(1−sin⁡B)(1−sin⁡C)(1 - \sin A)(1 - \sin B)(1 - \sin C),

we have, (1−sin⁡2A)(1−sin⁡2B)(1−sin⁡2C)(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)^{2}(1 - \sin B)^{2}(1 - \sin C)^{2}

⇒ (1−sin⁡A)(1−sin⁡B)(1−sin⁡C)=±cos⁡Acos⁡Bcos⁡C(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)=±cos⁡Acos⁡Bcos⁡C(1 + \sin A)(1 + \sin B)(1 + \sin C) = \pm \cos A\cos B\cos C.