Respuesta :

[tex]\bf \cfrac{cos^3(\theta)+sin^3(\theta)}{sin(\theta)cos(\theta)+sin^2(\theta)}=csc(\theta)-cos(\theta)\\\\ -----------------------------\\\\ \textit{let us do the left-hand-side} \\\\ recall\qquad \textit{difference of cubes} \\ \quad \\ a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad (a+b)(a^2-ab+b^2)= a^3+b^3 \\\\[/tex]

[tex]\bf -----------------------------\\\\ thus \\\\ \cfrac{[cos(\theta)+sin(\theta)][cos^2(\theta)-cos(\theta)sin(\theta)+sin^2(\theta)]}{sin(\theta)cos(\theta)+sin^2(\theta)} \\\\\\ \textit{let us take common factor on the denominator} \\\\\\ \cfrac{[\boxed{cos(\theta)+sin(\theta)}][cos^2(\theta)-cos(\theta)sin(\theta)+sin^2(\theta)]}{sin(\theta)[\boxed{cos(\theta)+sin(\theta)}]} \\\\\\\\ \cfrac{cos^2(\theta)-cos(\theta)sin(\theta)+sin^2(\theta)}{sin(\theta)}\\\\ -----------------------------\\\\ [/tex]

[tex]\bf \textit{now, recall your pythagorean identities}\qquad sin^2(\theta)+cos^2(\theta)=1\\\\ -----------------------------\\\\ \cfrac{\boxed{cos^2(\theta)+sin^2(\theta)}-cos(\theta)sin(\theta)}{sin(\theta)}\implies \cfrac{1-cos(\theta)sin(\theta)}{sin(\theta)} \\\\\\ \textit{now, distributing the denominator} \\\\\\ \cfrac{1}{sin(\theta)}-\cfrac{cos(\theta)\boxed{sin(\theta)}}{\boxed{sin(\theta)}}\implies \implies csc(\theta)-cos(\theta)[/tex]
(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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