Question

sin 6 theta + cos 6 theta + 3 sin square theta into cos square theta is equal to 1

Answer 1

Hope this answer is helpful..

Answer 2

$LHS \red{= sin^{6} \theta + cos^{6} \theta + 3sin^{2} \theta cos^{2} \theta }$

$= (sin^{2} \theta)^{3} + (cos^{2} \theta)^{3} + 3sin^{2} \theta cos^{2} \theta$

$=[ (sin^{2}\theta + cos^{2} \theta)^{3} - 3sin^{2}\theta cos^{2} \theta ( sin^{2}\theta cos^{2} \theta ] + 3sin^{2} \theta cos^{2} \theta$

$\boxed{\pink {Since , a^{3} + b^{3} = (a+b)^{3} - 3ab(a+b) }}$

$= 1 - 3sin^{2}\theta cos^{2} \theta \times 1+ 3sin^{2} \theta cos^{2} \theta$

$\boxed{\pink {Since , sin^{2}\theta + cos^{2} \theta =1 }}$

$\green {= 1 } \\= RHS$

$Hence \:proved$

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