if tan theta + sin theta equal m , tan theta minus sin theta equal n and then show that m² + n² equal 4√mn
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Answered by
4
Answer:
Tanθ+sinθ=m and tanθ-sinθ=n
∴, m²-n²
=(m+n)(m-n)
=(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)
=(2tanθ)(2sinθ)
=4tanθsinθ
4√mn
=4√(tanθ+sinθ)(tanθ-sinθ)
=4√(tan²θ-sin²θ)
=4√{(sin²θ/cos²θ)-sin²θ}
=4√sin²θ{(1/cos²θ)-1}
=4sinθ√{(1-cos²θ)/cos²θ}
=4sinθ√(sin²θ/cos²θ)
=4sinθ√tan²θ
=4sinθtanθ
∴, LHS=RHS Hence Proved
Answered by
1
Answer:
Tanθ+sinθ=m and tanθ-sinθ=n
∴, m²-n²
=(m+n)(m-n)
=(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)
=(2tanθ)(2sinθ)
=4tanθsinθ
4√mn
=4√(tanθ+sinθ)(tanθ-sinθ)
=4√(tan²θ-sin²θ)
=4√{(sin²θ/cos²θ)-sin²θ}
=4√sin²θ{(1/cos²θ)-1}
=4sinθ√{(1-cos²θ)/cos²θ}
=4sinθ√(sin²θ/cos²θ)
=4sinθ√tan²θ
=4sinθtanθ
∴, LHS=RHS Hence Proved
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