Question

If tanA-sinA=m and tanA-sinA=n, prove that m^2-n^2=4√mn.​

Answer 1

Explanation:

your question is wrong.

tanθ+sinθ=m

tanθ+sinθ=mtanθ-sinθ=n

tanθ+sinθ=mtanθ-sinθ=n∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθ

tanθ+sinθ=mtanθ-sinθ=n∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθm-n=tanθ+sinθ-tanθ+sinθ=2sinθ

tanθ+sinθ=mtanθ-sinθ=n∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθm-n=tanθ+sinθ-tanθ+sinθ=2sinθmn=(tanθ+sinθ)(tanθ-sinθ)

tanθ+sinθ=mtanθ-sinθ=n∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθm-n=tanθ+sinθ-tanθ+sinθ=2sinθmn=(tanθ+sinθ)(tanθ-sinθ) =tan²θ-sin²θ

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

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)

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

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=4sinθtanθ

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.

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