2tantheta/1+tan^²theta+2tantheta/1-tan^²theta=2cos^²theta×tan2theta
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Answer:
Step-by-step explanation:
2Tanθ / 1 + Tan²θ + 2Tanθ/ 1 - Tan²θ
=> [2(Sinθ/Cosθ) / 1 + (Sin²θ/Cos²θ) ] + Tan2θ (∵2Tanθ/ 1 - Tan²θ = Tan2θ)
=> [2SinθCos²θ/Cosθ /Cos²θ + Sin² θ] + Tan2θ
=> Sin2θ + Tan2θ (∵ Sin2θ = 2SinθCosθ)
=> Sin2θ + Sin2θ/Cos2θ
=> Sin2θ[ 1 + 1/Cos2θ]
=> Sin2θ [ Cos2θ + 1 / Cos2θ]
=> Sin2θ/Cos2θ ( 1 + Cos2θ)
=> Tan2θ * 2Cos²θ (∵ Cos2θ = 2Cos²θ - 1)
= R.H.S
Hence proved
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