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Question

# if (a2−b2)sinθ+2abcosθ=a2+b2, then prove tanθ=a2−b22ab

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Solution

## According to question.......(a2−b2)sinθ+2abcosθ=a2+b2⇒(a2+b2)−(a2−b2)sinθ=2abcosθNow,squaringbothside,⇒((a2+b2)−(a2−b2)sinθ)2=(2abcosθ)2⇒(a2+b2)2+(a2−b2)2sin2θ−2(a2+b2)(a2−b2)sinθ=4a2b2cos2θ⇒(a2+b2)2+(a2−b2)2sin2θ−2(a2+b2)(a2−b2)sinθ−4a2b2(1−sin2θ)=0⇒(a2+b2)2+(a2−b2)2sin2θ−2(a2+b2)(a2−b2)sinθ−4a2b2+4a2b2sin2θ)=0⇒[(a2−b2)2+4a2b2]sin2θ−2(a2+b2)2(a2−b2)sinθ+[(a2+b2)2−4a2b2]=0⇒(a2+b2)2sin2θ−2(a2+b2)2(a2−b2)sinθ+(a2−b2)2=0⇒[(a2+b2)2sinθ−(a2−b2)]2=0⇒(a2+b2)sinθ−(a2−b2)=0⇒sinθ=(a2−b2)(a2+b2)⇒cosecθ=a2+b2a2−b2Now,cot2θ=cosec2θ−1⇒cot2θ=(a2+b2a2−b2)2−1⇒cot2θ=(a2+b2)2−(a2−b2)2(a2−b2)2=(a2+b2−a2+b2)(a2+b2+a2−b2)(a2−b2)2⇒cot2θ=(2a2)(2b2)(a2−b2)2=4a2b2(a2−b2)2⇒cotθ=√4a2b2(a2−b2)2=2aba2+b2⇒tanθ=a2+b22ab∣∣cotθ=1tanθsothatweprove:tanθ=a2+b22ab

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