Question

prove that 2(Sin⁶θ+Cos⁶θ)-3(Sin⁴θ+Cos⁴θ)+1 = 0. ​

Answer 1

Given :-

• 2(Sin⁶θ+Cos⁶θ)-3(Sin⁴θ+Cos⁴θ)+1

To Prove :-

• LHS = 0

Step-by-step explanation:

$\sf \: LHS = 2( \sin^{6} \theta + \cos^{6} \theta ) - 3( \sin^{4} \theta + { \cos^{4} \theta}) + 1$

$\implies \sf \: 2 [( \sin^{2} \theta)^{3} + ( \cos^{2} \theta)^{3} ] - 3[( \sin^{2} \theta)^{2} + ( \cos^{2} \theta)^{2} ] + 1$

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

$\implies \sf \: 2[( 1)^{3} - 3 \sin^{2} \theta \cos^{2} \theta \times 1] - 3[(1)^{2} - 2\sin^{2} \theta \cos^{2} \theta] + 1$

$\implies \sf \: 2(1 - 3\sin^{2} \theta \cos^{2} \theta) - 3(1 - 2\sin^{2} \theta \cos^{2} \theta) + 1$

$\implies \sf \: 2 - 6\sin^{2} \theta \cos^{2} \theta - 3 + 6\sin^{2} \theta \cos^{2} \theta + 1$

$\implies \sf \: 2 - 3 + 1$

$\implies \boxed{ \red{ \bold{ \sf \: 0 \: }}} = \sf \: RHS$

Hence, Proved!

Answer 2

Answer:

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