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Prove that (m2 -n2)4/mn |
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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=(m+n)(m-n)=2tanθ.2sinθ=4sinθtanθ4√mn=4√(tan²θ-sin²θ)=4√(sin²θ/cos²θ-sin²θ)=4√sin²θ(1/cos²θ-1)=4sinθ√(1-cos²θ)/cos²θ=4sinθ/cosθ√sin²θ [∵,sin²θ+cos²θ=1]=4sinθtanθ∴, LHS=RHS (proved) please like the solution 👍 ✔️👍 |
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