Answer: We will start by using the identity cos² A + sin² A = 1 to replace sin² A with cos² A - 1 in the expression 1 - 2sin A sin B sin C:
1 - 2sin A sin B sin C = 1 - 2sin A (cos B cos C - sin B sin C) [Using the product-to-sum formula for sin (B + C)]
= 1 - 2sin A cos B cos C + 2sin A sin B sin C
Now, we will use the identity cos (180° - x) = -cos x to replace cos C with -cos (A + B):
cos² A + cos² B - cos² C = cos² A + cos² B - cos² (A + B)
= cos² A + cos² B - (cos² A cos² B - 2cos A cos B sin A sin B)
= cos² A + cos² B - cos² A cos² B + 2cos A cos B sin A sin B
= (cos² A)(1 - cos² B) + (cos² B)(1 - cos² A) + 2cos A cos B sin A sin B
= cos² A + cos² B - cos² A cos² B + 2cos A cos B sin A sin B
Now, we will use the identity sin (A + B) = sin A cos B + cos A sin B to rewrite the last term:
cos² A + cos² B - cos² A cos² B + 2cos A cos B sin A sin B
= cos² A + cos² B - cos² A cos² B + 2sin A sin B cos A cos B
= (cos² A)(1 - cos² B) + (cos² B)(1 - cos² A) + 2sin A sin B cos A cos B
= (cos² A + cos² B - 1) + (1 - cos² A)(1 - cos² B) + 2sin A sin B cos A cos B
= 2sin² A sin² B + 2sin A sin B cos A cos B
= 2sin A sin B (sin A cos B + cos A sin B)
= 2sin A sin B sin (A + B)
= 2sin A sin B sin (180° - C) [Using A + B + C = 180°]
= -2sin A sin B sin C
Substituting this expression back into the original equation, we get:
cos² A + cos² B - cos² C = 1 - 2sin A sin B sin C
Therefore, the expression is proved.
Explanation: