if tan θ+ sin θ=m and tan θ-sin θ=n,show that
m²-n²=4√mn
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tanθ+sinθ=m
tanθ-sinθ=n
∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθ
m-n=tanθ+sinθ-tanθ+sinθ=2sinθ
mn=(tanθ+sinθ)(tanθ-sinθ)
=tan²θ-sin²θ
∴, m²-n²
=(m+n)(m-n)
=2tanθ.2sinθ
=4sinθtanθ
4√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)
already answered in brainly
refer:-https://brainly.in/question/732979
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