Question

(sinA+cosA)(1-sinA.cosA)=sin³A+cos³A​

Answer 1

Step-by-step explanation:

a³+b³=(a+b)(a²-ab+b²)

Sin³A+cos³A=(sinA+cosA)(sin²A-sinAcosA+cos²A)

=(sinA+cosA)(1-sinAcosA)

Answer 2

LHS:

(sinA+cosA)(1-sinA.cosA)

RHS:

sin³A+cos³A

To prove:

LHS = RHS

By taking RHS..

sin³A+cos³A

=> (sinA+cosA)(sin²A-sinAcosA+cos²A) [using formula = a³+b³ =(a+b)(a²-ab+b²)]

=>(sinA+cosA)(sin²A+cos²A-sinAcosA)

=>(sinA+cosA)(1-sinAcosA) [using identity = sin²ø+cos²ø = 1]

Therefore,

LHS= RHS

Hence, proved✔✔

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