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24.lftane + sinθm and tanA-sinen, show that (m2-n2)2-16mn |
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Answer» Tanθ+sinθ=mtanθ-sinθ=n∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθm-n=tanθ+sinθ-tanθ+sinθ=2sinθmn=(tanθ+sinθ)(tanθ-sinθ) =tan²θ-sin²θ∴(m²-n²)²=((m+n)(m-n))²=(2tanθ.2sinθ)²=16sin²θtan²θ 16mn=16(tan²θ-sin²θ)=16(sin²θ/cos²θ-sin²θ)=16sin²θ(1/cos²θ-1)=16sin²θ(1-cos²θ)/cos²θ=4sin²θ/cos²θ.sin²θ [∵,sin²θ+cos²θ=1]=16sin²θtan²θ Hence proved. |
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