To prove-->Sin^6€+Cos^6€+3Sin^2€.Cos^2€=1,
L.H.S-->
Sin^6€+Cos^6€+3sin^2€.Cos^2€,
=(sin^2)^3€+(Cos^2)^3€+3Sin^2€.Cos^2€,
=(sin^2)^3€+(cos^2€)^3€+3sin^2€.cos^2€(sin²€+cos²€)
=(sin^2€+cos^2€)^3,
=1,
LHS=RHS,
Hence , proved.

Proof:

To prove that sin^6 θ * cos^6 θ * 3sin^2 θ * cos^2 θ = 1, we will break down the expression and simplify it step by step.

Step 1: Simplify sin^6 θ * cos^6 θ

We can rewrite sin^6 θ as (sin^2 θ)^3 and cos^6 θ as (cos^2 θ)^3. Therefore, we have:

(sin^2 θ)^3 * (cos^2 θ)^3

Step 2: Apply the identity sin^2 θ + cos^2 θ = 1

Using the identity sin^2 θ + cos^2 θ = 1, we can rewrite the expression as:

[(1 - cos^2 θ)^2] * [(1 - sin^2 θ)^2]

Step 3: Expand the expression

Expanding the expression, we get:

(1 - 2cos^2 θ + cos^4 θ) * (1 - 2sin^2 θ + sin^4 θ)

Step 4: Simplify the expression

Multiplying the terms, we have:

1 - 2cos^2 θ + cos^4 θ - 2sin^2 θ + 4sin^2 θ * cos^2 θ - 2sin^4 θ + cos^4 θ - 2cos^2 θ * sin^2 θ + sin^4 θ

Simplifying further, we get:

1 - 2cos^2 θ - 2sin^2 θ + cos^4 θ + 4sin^2 θ * cos^2 θ - 2sin^4 θ

Using the identity sin^2 θ + cos^2 θ = 1, we can simplify the expression to:

1 - 2 + cos^4 θ + 4sin^2 θ * cos^2 θ - 2sin^4 θ

Simplifying further, we have:

-cos^4 θ + 4sin^2 θ * cos^2 θ - 2sin^4 θ - 1

Step 5: Rearrange the terms

Rearranging the terms, we get:

-1 + 4sin^2 θ * cos^2 θ - cos^4 θ - 2sin^4 θ

Step 6: Apply the identity sin^2 θ * cos^2 θ = 1/4 * sin^2(2θ)

Using the identity sin^2 θ * cos^2 θ = 1/4 * sin^2(2θ), we can rewrite the expression as:

-1 + 4 * (1/4) * sin^2(2θ) - cos^4 θ - 2sin^4 θ

Simplifying further, we have:

-1 + sin^2(2θ) - cos^4 θ - 2sin^4 θ

Step 7: