Question

please give me an written explanation
please
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if i like your answer​

Answer 1

To prove--->

(1 - 4Sin²A )³ + (3 - 4Cos²A )³

---------------------------------------

(√3 + 2CosA )³ ( 1 + 2SinA )³

Proof ---> First we find value of ( 1 - 4Sin²A )

We have an identity, Sin²A = 1 - Cos²A , we get

1 - 4 Sin²A = 1 - 4 ( 1 - Cos²A )

= 1 - 4 + 4 Cos²A

= - 3 + 4 Cos²A

= - ( 3 - 4 Cos²A )

Now solving numerator first

( 1 - 4Sin²A )³ + ( 3 - 4Cos²A )³

Putting 1 - 4Sin²A = - ( 3 - 4Cos²A ) , we get

= { - ( 3 - 4Cos²A ) }³ + ( 3 - 4Cos²A )³

= - ( 3 - 4Cos²A )³ + ( 3 - 4Cos²A )³

= 0

Now, LHS

( 1 - 4Sin²A )³ + ( 3 - 4 Cos²A )

= ------------------------------------------

( √3 + 2 CosA ) ( 1 + 2 SinA )³

Putting value of numerator from above , we get

0

= -----------------------------------------

( √3 + 2 CosA )³ ( 1 + 2SinA )³

= 0 = RHS