Question

If cos A + cos B = cos A ×cos B-sin A ×sin B then prove Sin A +sin B = sin A×cos B+cos A×sin B

Answer 1

Step-by-step explanation:

Multiplying both sides by 2 in given equality cosAcosB+sinAsinBsinC=1, we get

2cosAcosB+2sinAsinBsinC=2 or

2cosAcosB+2sinAsinBsinC=(sin2A+cos2A)+(sin2B+cos2B)

or (cos2A+cos2B−2cosAcosB)+(sin2A+sin2B−2sinAsinB)+2sinAsinB−2sinAsinBsinC=0 or

or (cosA−cosB)2+(sinA−sinB)2+2sinAsinB(1−sinC)=0

Note that all three terms contained above are positive, as while first two terms are squares and hence positive, third term is positive as sine of angles A and B is positive (as they are less than π being angles of a triangle) and 

Answer 2

Answer:

LHS = RHS

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