Question

Sin6thita+cos6thita=1-3sin2thita×cos2thita

Answer 1

$\bf{hey !!!}$

from LHS

sin^6 ¢ + cos^6¢

= ( sin²¢ )³ + (cos²¢)³

°•° we know that a³ + b³ = ( a + b ) (a² + b² - ab)

using this .

we get

= (sin²¢ + cos²¢ ) ( sin⁴¢ + cos⁴¢ - sin²¢ cos²¢ )

= 1 { ( sin²¢ )² + ( cos²¢ )² - sin²¢ cos²¢ }

[using rule a² + b² = ( a + b)²-2ab]

= { (sin²¢ + cos²¢ )² - 2sin²¢ cos²¢ - sin²¢ cos²¢

= 1 - 3sin²¢ cos²¢ prooved ♻

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