Question

sin ³a +cos ³a/sina + cos a + sina cosa =1​

Answer 1

To Prove : [ { sin³A + cos³A } / { sinA + cosA } ] + cosA sinA = 1

Proof :

From the properties of expansion we know that a³ + b³, when factorised, is ( a + b)( a² + b² - ab) .

Therefore,

sin³A + cos³A will be ( sinA + cosA )( cos²A + sin²A - cosAsinA )

Thus,

= > [ { ( sinA + cosA )( sin²A + cos²A - sinAcosA) } / { sinA + cosA } ] + sinAcosA

Removing ( sinA + cosA) from numerator and denominator both.

From the properties of trigonometry, we know : sin²A + cos²A

= > [ 1 - sinA cosA ] + sinAcosA

= > 1 - sinAcosA + sinAcosA

= > 1

Hence proved.

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