Please solve this

Asked by  | 5th Oct, 2008, 09:03: PM

Expert Answer:

 (a2-b2)sinθ + 2abcosθ = a2+b2        divided both side by {(a2-b2)2+(2ab)2 }  =  (a2+b2)= a2+b2

we get  (a2-b2)sinθ/(a2+b2) + 2abcosθ(a2+b2)  = 1    let cos k = (a2-b2)/(a2+b2)  ====> sin k =2ab/(a2+b2)    this imply tan k =2ab/(a2 - b2)  

we get  cos k sinθ + sink cos θ = 1=====>  sin(θ+k) = 1===> θ + k = 900 ====> θ = 900- k

so tan θ =  tan(900- k ) = cot k  = (a2 - b2) / 2ab

tan θ = (a2 - b2) / 2ab

Answered by  | 16th Dec, 2008, 09:52: PM

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