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3 years ago

Prove. sin³A-cos³A/sinA-cosA = 1+sinAcosA

Mathematics

1 answer:

katovenus [111]3 years ago

7 0

Answer:

$\frac{ \sin {}^{3} A - \cos {}^{3} A }{ \sin A - \cos A} \\ \\ { \sf{ = \frac{ {( \sin A - \cos A)}^{3} + 3 \sin A \cos A( \sin A - \cos A)}{ \sin A - \cos A} }} \\ \\ = { {( \sin A - \cos A)}^{2} + 3 \sin A \cos A} \\ { \sf{ = ( \sin {}^{2} A + \cos {}^{2} A) - 2 \sin A \cos A + 3\sin A \cos A}} \\ = { \sf{1 +\sin A \cos A }}$

#hence proved

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3 years ago

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Answer:

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Step-by-step explanation:

(a)

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$=10979\pm 1.729\times \frac{1000}{\sqrt{20}}\\=10979\pm4.47\\ =(10974.53, 10983.47)$

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(b)

The margin of error is provided as:

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3 years ago

Prove. sin³A-cos³A/sinA-cosA = 1+sinAcosA​

Prove. sin³A-cos³A/sinA-cosA = 1+sinAcosA