Question

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Answer 1

Hey mate,.

2(sin6theta+cos6theta)-3(sin⁴+cos⁴)+1=0

2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1

= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1

= 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1

=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1

=2-6sin²θcos²θ-3+6sin²θcos²θ+1

=-1+1

=0 (Proved)

Hope it will help you.

✌️sai

Answer 2

Hii !

Solution

From LHS

= 2(sin^6¢ + cos^6 ) - 3(sin⁴¢ + cos⁴¢ ) + 1

= 2{( sin²¢ )³+( cos²¢ )³} -3 {(sin²¢ )² + (cos²¢)² }+ 1

We know that , a³ + b³ = (a + b) (a² + b² - ab) , and also know (a²+b²) = (a + b)²-2a²b²

using , this identity

since,

= 2 (sin²¢ +cos²¢) ( sin⁴¢ + cos⁴¢- sin²¢ * cos²¢ ) - 3 {(sin²¢ + cos²¢ )² - 2sin²¢ cos²¢} + 1

= 2{(sin²¢)²+(cos²¢)² - sin²¢ cos²¢ -3 (1 - 2sin²¢ . cos²¢ ) + 1

= 2 {(sin²¢ + cos²¢ )² - 2sin²¢ cos²¢ - sin²¢ cos²¢ } - 3(1 - 2sin²¢.cos²¢) + 1

= 2 ( 1 - 3sin²¢ .cos²¢ ) } - 3 (1 - 2sin²¢ .cos²¢) + 1

= 2 - 6sin²¢ . cos²¢ - 3 + 6sin²¢ .cos²¢ + 1

= 3 - 3 = 0 LHS = RHS prooved

_____________________________

Hope it helps you !!

@Raj❤