Math, asked by sumidigital143, 10 months ago

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sin3 A +cos3 A/sin A + cos A
sin3 A - cos3 A/sin A - cos A=2​

Answers

Answered by harekrushnanayak
10

We have to prove that

(sin³A+cos³A/sinA+cosA) + (sin³A-cos³A/sinA-cosA)=2

LHS = (sin³A+cos³A/sinA+cosA) + (sin³A-cos³A/sinA-cosA)

= {(sinA+cosA)(sin²A+cos²A-sinA.cosA)}/(sinA+cosA) + {(sinA-cosA)(sin²A+cos²A+sinA.cosA)}/(sinA-cosA)

= (sin²A+cos² on Brainly.in - https://brainly.in/question/29033

We have to prove that

(sin³A+cos³A/sinA+cosA) + (sin³A-cos³A/sinA-cosA)=2

LHS = (sin³A+cos³A/sinA+cosA) + (sin³A-cos³A/sinA-cosA)

= {(sinA+cosA)(sin²A+cos²A-sinA.cosA)}/(sinA+cosA) + {(sinA-cosA)(sin²A+cos²A+sinA.cosA)}/(sinA-cosA)

= (sin²A+cos²A-sinA.cosA) + (sin²A+cos²A+sinA.cosA)

= 1-sinA.cosA+1+sinA.cosA

= 1+1

= 2 = RHS [pro

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