Math

Cos square theta divided by 1-tan theta +sin cube theta divided by sin theta -cos theta = 1+sin theta *cos theta

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  1. cos^2 θ /(1 - tanθ) + sin^3 θ/(sinθ - cosθ) = 1 + sinθ cosθ
    LS = cos^2 θ /(1 - sinθ/cosθ) + sin^3 θ/(sinθ - cosθ)
    = cos^2 θ /((cosθ - sinθ)/cosθ) + sin^3 θ/(sinθ - cosθ)
    = cos^3 θ /(cosθ - sinθ) - sin^3 θ/(cosθ - sinθ)
    = (cos^3 θ - sin^3 θ)/(cosθ - sinθ) <---- I see the difference of cubes
    = (cosθ - sinθ)(cos^2 θ + sinθcosθ + sin^2 θ)/(cosθ - sinθ)
    = (cos^2 θ + sinθcosθ + sin^2 θ)
    = 1 + sinθcosθ
    = RS

    wheeww!!

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  2. If your question mean.

    Prove:

    cos² ( θ ) / [ 1 - tan ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] = 1 + sin ( θ ) ∙ cos ( θ )

    then

    cos² ( θ ) / [ cos ( θ ) / cos ( θ ) - sin ( θ ) / cos ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

    cos² ( θ ) / [ cos ( θ ) - sin ( θ ) ] / cos ( θ ) + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

    cos³ ( θ ) / [ cos ( θ ) - sin ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

    - cos³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

    [ sin³ ( θ ) - cos³ ( θ ) ] / [ sin ( θ ) - cos ( θ ) ] = 1 + sin ( θ ) ∙ cos ( θ ) =

    ___________________________
    Since:
    a³ - b³ = ( a - b ) ∙ ( a² + a ∙ b + b² )
    ___________________________

    [ sin ( θ ) - cos ( θ ) ] ∙ [ sin² ( θ ) + sin ( θ ) ∙ cos ( θ ) + cos ( θ )² ] / [ sin ( θ ) - cos ( θ ) ] =

    sin² ( θ ) + sin ( θ ) ∙ cos ( θ ) + cos ( θ )² =

    sin² ( θ ) + cos ( θ )² + sin ( θ ) ∙ cos ( θ ) =

    1 + sin ( θ ) ∙ cos ( θ )

    By the way:

    1 + sin ( θ ) ∙ cos ( θ ) = 1 + ( 1 / 2 ) sin ( 2 θ )

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