Question

prove that:

(sin3x+cos3x)/(sinx+cosx)+(sin3x-cos3x)/(sinx-cosx)=2​

Answer 1

Appropriate Question :-

Prove that :-

$\sf \: \dfrac{ {sin}^{3}x + {cos}^{3}x}{sinx + cosx} + \dfrac{ {sin}^{3} x - {cos}^{3} x}{sinx - cosx} = 2$

Consider LHS

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

We know,

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

and

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

So, using these Identities, we get

$\sf=\dfrac{\cancel{(sinx+cosx)}( {sin}^{2}x + {cos}^{2}x-sinxcosx)}{\cancel{sinx + cosx}}+\dfrac{\cancel{(sinx - cosx)}( {sin}^{2}x + {cos}^{2}x + sinxcosx)}{\cancel{sinx - cosx}}$

$\sf =( {sin}^{2}x + {cos}^{2}x - sinxcosx) + ( {sin}^{2}x + {cos}^{2}x + sinxcosx)$

We know,

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

So, using this, we get

$\sf   =1 - \cancel{sinxcosx} + 1 + \cancel{sinxcosx}$

$\sf \:  =  \: 2$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \: \dfrac{ {sin}^{3}x + {cos}^{3}x}{sinx + cosx} + \dfrac{ {sin}^{3} x - {cos}^{3} x}{sinx - cosx} = 2}}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1