Question

(sin³A + cos³ A/sin A + cos A) + sinA.cosA = 1​

Answer 1

Given :

(sin³A + cos³ A/sin A + cos A) + sinA.cosA

To prove :

(sin³A + cos³ A/sin A + cos A) + sinA.cosA = 1

Solution :

=> (sin³A + cos³ A/sin A + cos A) + sinA.cosA

=> sinA + cosA (sin²A + cos²A - sinAcosA) + sinAcosA/sinA + cosA

=> sin²A + cos²A - sinAcosA + sinA. cosA

=> sin²A + sin²A

=> 1

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

${sin}^{2} A + {cos}^{2} A = 1$

2.

${a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$

TO PROVE

$\displaystyle \frac{sin³A + cos³ A }{sin A + cos A} + sin A \: cos A = 1$

PROOF

$\displaystyle \frac{sin³A + cos³ A }{sin A + cos A} + sin A \: cos A$

$= \displaystyle \frac{(sinA + cosA)({sin}^{2} A - sinA \: cosA + {cos}^{2} A )}{sin A + cos A} + sin A \: cos A$

$= \displaystyle \frac{(sinA + cosA)({sin}^{2} A + {cos}^{2} A - sinA \: cosA + )}{sin A + cos A} + sin A \: cos A = 1$

$= \displaystyle \: ({sin}^{2} A + {cos}^{2} A - sinA \: cosA )+ sin A \: cos A$

$= \displaystyle \: 1- sinA \: cosA + sin A \: cos A$

$= 1$

Hence proved