Question

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IF sin2a+sin2b=sin2c prove that either a=90° and b=90°.



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Answer 1

Answer:

A+B+C=180

sin2A+sin2B=sin2C

2sin(2A+2B)/2cos(2A−2B)/2=sin2C

2sin(A+B)cos(A−B)=sin2C

2sin(180−C)cos(A−B)=sin2C

2sinCcos(A−B)=2sinCcosC

cos(A−B)=cosC

A−B=C

A=B+C

A=180−A

A=90

OR

cos(A−B)=cos(B−A)=cosC

B−A=C

B=A+C

B=180−B

B=90

Answer 2

Step-by-step explanation:

2sin(2A+2B)/2cos(2A−2B)/2=sin2C

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C2sin(180−C)cos(A−B)=sin2C

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C2sin(180−C)cos(A−B)=sin2C2sinCcos(A−B)=2sinCcosC

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C2sin(180−C)cos(A−B)=sin2C2sinCcos(A−B)=2sinCcosCcos(A−B)=cosC

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C2sin(180−C)cos(A−B)=sin2C2sinCcos(A−B)=2sinCcosCcos(A−B)=cosCA−B=C

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C2sin(180−C)cos(A−B)=sin2C2sinCcos(A−B)=2sinCcosCcos(A−B)=cosCA−B=CA=B+C

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C2sin(180−C)cos(A−B)=sin2C2sinCcos(A−B)=2sinCcosCcos(A−B)=cosCA−B=CA=B+CA=180−A

2sin(2A+2B)/2cos(2A−2B)/2=sin2C2sin(A+B)cos(A−B)=sin2C2sin(180−C)cos(A−B)=sin2C2sinCcos(A−B)=2sinCcosCcos(A−B)=cosCA−B=CA=B+CA=180−AA=90

And

cos(A−B)=cos(B−A)=cosC

cos(A−B)=cos(B−A)=cosCB−A=C

cos(A−B)=cos(B−A)=cosCB−A=CB=A+C

cos(A−B)=cos(B−A)=cosCB−A=CB=A+CB=180−B

cos(A−B)=cos(B−A)=cosCB−A=CB=A+CB=180−BB=90