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Prove that : a^2sin2c + c^2 sin2A = 4∆ By using tangent rule and half angels rule

Prove  that  :    a^2sin2c  +  c^2 sin2A  =  4∆
      By using tangent rule and half angels rule
 

Grade:11

1 Answers

Arun
25763 Points
3 years ago
b2sin2C + c2sin2B    = b2(2sinC cosC) + c2(2sinB cosB) Since, sinC/c =sinB/b = sinA/a = R   So,  = b2(2Rc cosC) + c2(2RbcosB)   =b2[2Rc ×{b2+a2 - c2}/(2ab)]+c2[2Rb×{a2 + c2-b2}/(2ac)]   =2Rbc [ {b2+a2 - c2}/(2a) + {a2 + c2-b2}/(2a)]   =2Rbc [ a]   = 2bc(Ra)   =2bcsinA = 4 * Area of triangle ABC

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