Ctheta%29%3Dcsc%5Ctheta-cos%5Ctheta" id="TexFormula1" title=" (cos^3 \theta +sin^3\theta) /(sin\theta cos\theta+sin^2\theta)=csc\theta-cos\theta" alt=" (cos^3 \theta +sin^3\theta) /(sin\theta cos\theta+sin^2\theta)=csc\theta-cos\theta" align="absmiddle" class="latex-formula">
2 answers:
(cos^3Ф+sin^3Ф)/sinФcosФ+sin^2Ф=cosecФ-cosФ
lets take left hand side first
(cos^3Ф+sin^3Ф)/sinФcosФ+sin^2Ф
(A^3+B^3)=(A+B)(A^2-AB+B^2)
(cosФ+sinФ)(cos^2Ф-cosФsinФ+sin^2Ф)/sinФcosФ+sin^2Ф
(sin^2x+cos^2x=1)
(cosФ+sinФ)(1+sinФcos)/sinФcosФ+sin^2Ф
(cosФ+sinФ)1/2*(2-2sinФcosФ) /sinФcosФ+sin^2Ф
1/2*(cosФ+sin)(2-sin2Ф)/sinФcosФ+sin^2Ф
1/2*(2-sin2Ф)/sinФ
now lets take RHS
cosecФ-cosФ=1/sinФ-cosФ
(1-sinФcosФ)/sinФ
1/2*(2-2sinФcosФ)/sinФ
1/2*(2-2sin2Ф)/sinФ
Hence proved
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