Question

Pls help me prove this Trigonometry Sum...proper Answer will be marked brainliest ​

Answer 1

Given Question :-

Prove that

$\sf \: \dfrac{ {cos}^{3}A - {sin}^{3}A}{cosA - sinA} - \dfrac{ {cos}^{3}A + {sin}^{3}A}{cosA + sinA} = 2sinA \: cosA$

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

Consider LHS

$\rm :\longmapsto\: \sf \: \dfrac{ {cos}^{3}A - {sin}^{3}A}{cosA - sinA} - \dfrac{ {cos}^{3}A + {sin}^{3}A}{cosA + sinA}$

We know that,

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

and

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

So, using these Identities, we get

$\sf \:  =\dfrac{\cancel{(cosA - sinA)}({cos}^{2}A + {sin}^{2}A + cosAsinA) }{\cancel{cosA - sinA}} \: - \: \\ \sf \: \dfrac{\cancel{(cosA + sinA)}( {cos}^{2} A + {sin}^{2} A - cosAsinA)}{\cancel{cosA + sinA}}$

We know,

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

So, using this, we get

$\sf \:  =  \: (1 + sinA \: cosA) - (1 - sinA \: cosA)$

$\sf \:  =  \: \cancel{1} + sinA \: cosA - \cancel{1} + sinA \: cosA$

$\sf \:  =  \: 2 \: sinA \: cosA$

Hence, Proved

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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