Question

can anyone help me in this question ​

Answer 1

Answer:

Plz mark me as brainliest

Step-by-step explanation:

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)

Answer 2

Answer:

Here you go ( I hope its correct )

Step-by-step explanation:

2( sin⁶θ + cos⁶θ ) - 3( sin⁴θ + cos⁴θ ) + 1 = 0

Solving LHS :

2[ (sin²θ)³ + (cos²θ)³ + 3(sin²θ)(cos²θ)(sin²θ + cos²θ) - 3(sin²θ)(cos²θ)(sin²θ + cos²θ) ] - 3[ (sin²θ)² + (cos²θ)² + 2(sin²θ)(cos²θ) - 2(sin²θ)(cos²θ) ] - 1 = 0  

   

{ Since

0 =  3(sin²θ)(cos²θ)(sin²θ + cos²θ) - 3(sin²θ)(cos²θ)(sin²θ + cos²θ) ,

0 = 2(sin²θ)(cos²θ) - 2(sin²θ)(cos²θ) }

⇒2[ (sin²θ + cos²θ)³  - 3(sin²θ)(cos²θ)(sin²θ + cos²θ) ] - 3 [ (sin²θ + cos²θ)² - 2(sin²θ)(cos²θ) ] - 1

{ Since

( sin²θ + cos²θ)³ = (sin²θ)³ + (cos²θ)³ + 3(sin²θ)(cos²θ)(sin²θ + cos²θ),

( sin²θ + cos²θ )² = (sin²θ)² + (cos²θ)² + 2(sin²θ)(cos²θ) }

⇒2 [ (1)³ - 3(sin²θ)(cos²θ)(1) ] - 3[ (1)² - 2(sin²θ)(cos²θ) ] - 1

{Since sin²θ + cos²θ = 1}

⇒2[ 1 - 3(sin²θ)(cos²θ) ] - 3 [ 1 - 2(sin²θ)(cos²θ) ] - 1

⇒2 - 6(sin²θ)(cos²θ) - 3 + 6(sin²θ)(cos²θ) - 1

⇒-1 + 1

= 0

= RHS

HENCE PROVED

HOPE IT HELPS

(Note : This was very difficult to write so i have tried to space it out but editing i perfectly is almost impossible so there might be some mistakes, although I have done my best)