Question

Prove that, tan(A + B) - tan(A - B) = sin 2 B÷ (cos²B- sin²A)​

Answer 1

Answer:

tan(A+B)−tan(A−B)

=tanA+tanB1−tanAtanB−tanA−tanB1+tanAtanB

=(tanA+tanB)(1+tanAtanB)−(tanA−tanB)(1−tanAtanB)(1−tanAtanB)(1+tanAtanB)

=2tanB+2tan2AtanB(1−tan2Atan2B)⋅cos2Acos2Bcos2Acos2B

=2sinBcos2AcosB+2sin2AsinBcosBcos2Acos2B−sin2Asin2B

=2sinBcosB(cos2A+sin2A)(1−sin2A)cos2B−sin2A(1−cos2B)

=sin(2B)(1)cos2B−sin2Acos2B−sin2A+sin2Acos2B

=sin(2B)cos2B−sin2A

Answer 2

your question is wrong,,

there should be sin2B/cos(A+B). cos(A-B)