CBSE Class 10 Answered
Please solve this
Asked by | 05 Oct, 2008, 09:03: PM
Expert Answer
(a2-b2)sinθ + 2abcosθ = a2+b2 divided both side by {(a2-b2)2+(2ab)2 } = (a2+b2)2 = 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 | 16 Dec, 2008, 09:52: PM
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