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Rajan@2021
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If (tanθ+sinθ)=m and (tanθ-sinθ)=n, prove that (m²-n²)² = 16mn

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This is the basic and conceptual question from trigonometry, topic –  trigonometric identities in which we have given that (tanθ+sinθ)=m and also (tanθ-sinθ)=n, and we need to  prove that (m²-n²)² = 16mn

RS Aggarwal, Class 10, chapter 13B, question no 7

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  1. (tan θ + sin θ) = m

    and (tan θ − sin θ) = n

    To Prove: (m² − n²)² = 16mn

    L.H.S. = (m² − n²)²

    = [(tan θ + sin θ)² – (tan θ − sin θ)²]² = (4tan θ sin θ)²

    = 16 tan²θ sin²θ…….(1)

    R.H.S.

    = 16mn

    = 16(tan θ + sin θ)(tan θ − sin θ)

    = 16(tan²θ − sin²θ)

    = 16 [{sin²θ (1-cos²θ)/cos²θ]

    = 16 × sin²θ/cos²θ × (1-cos²θ)

    = 16 tan²θ sin²θ …(2)

    From (1) and (2) L.H.S. = R.H.S.

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