Question

prove the identify sin³+cos³/sin cos =1-sin cos​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \: \dfrac{ {sin}^{3}x + {cos}^{3} x }{sinx + cosx} = 1 - sinxcosx$

$\large\underline{\sf{Solution-}}$

Basic Identities Used :-

$\boxed{ \bf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)}}$

$\boxed{ \bf{ \: \: {sin}^{2} x + {cos}^{2} x = 1}}$

Let's solve the problem now!!

Consider,

$\rm :\longmapsto\:\dfrac{ {sin}^{3}x + {cos}^{3} x }{sinx + cosx}$

$\rm :\longmapsto\: = \: \dfrac{ \cancel{ (sinx + cosx)}({sin}^{2}x + {cos}^{2} x - sinx \: cosx) } {\cancel{sinx + cosx}}$

$\rm :\longmapsto\: = \: 1 - sinxcosx$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)