Math, asked by agraharisuman038, 8 months ago

prove that sin²Acos2B-sin²Bcos2A=cos²B-cos²A

Answers

Answered by BrainlyTornado
4

GIVEN:

  • sin² A cos 2B - sin² B cos 2A = cos² B - cos² A

TO PROVE:

  • sin² A cos 2B - sin² B cos 2A = cos² B - cos² A

FORMULAE:

cos 2x = cos² x - sin² x

sin² x = ( 1 - cos² x)

PROOF:

  • Let L.H.S = sin² A cos 2B - sin² B cos 2A

  • Let R.H.S = cos² B - cos² A

L.H.S

  • sin² A cos 2B - sin² B cos 2A

 \bold{ \large{ \boxed{ \sf {\cos 2x =  { \cos}^{2} x -  { \sin}^{2} x}}}}

  • sin² A( cos² B - sin² B) - sin² B(cos² B - sin² B)

  • sin² A cos² B - sin² A sin² B - sin² B cos² A + sin² B sin² A

  • sin² A cos² B - sin² B cos² A

 \bold{ \large{ \boxed{ \sf { { \sin}^{2} x = 1 -  { \cos}^{2} x}}}}

  • ( 1 - cos² A) cos² B - (1 - cos² B)cos² A

  • cos² B - cos² A cos² B - cos² A + cos² B cos² A

  • cos² B - cos² A

R.H.S = cos² B - cos² A

L.H.S = R.H.S

HENCE PROVED.

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