Prove the following trigonometry identities

1. (1 βˆ’ π‘π‘œπ‘ 2πœƒ)𝑐𝑠𝑐2πœƒ = 1
2.π‘ π‘’π‘πœƒβˆš(1 βˆ’ 𝑠𝑖𝑛2πœƒ) = 1 [5]
3. π‘π‘œπ‘ πœƒ
(1βˆ’π‘‘π‘Žπ‘›πœƒ)
+
π‘ π‘–π‘›πœƒ
1βˆ’π‘0π‘‘πœƒ = π‘ π‘–π‘›πœƒ + π‘π‘œπ‘ πœƒ

To prove the first trigonometry identity, we start with the left side of the equation:

(1 βˆ’ cos^2ΞΈ) sinΞΈ cos^2ΞΈ

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

1 βˆ’ sin^2ΞΈ

Substituting this back into the previous equation, we get:

(1 βˆ’ cos^2ΞΈ) sinΞΈ (1 βˆ’ sin^2ΞΈ)

Expanding this multiplication gives:

sinΞΈ βˆ’ sin^3ΞΈ

Now, using the identity sin^3ΞΈ = sinΞΈ(1 βˆ’ cos^2ΞΈ), which can be derived from sin^2ΞΈ = 1 βˆ’ cos^2ΞΈ, we can rewrite the previous equation as:

sinΞΈ βˆ’ sinΞΈ(1 βˆ’ cos^2ΞΈ)

Simplifying further gives:

sinΞΈ βˆ’ sinΞΈ + sinΞΈcos^2ΞΈ

Combining like terms, we obtain:

sinΞΈcos^2ΞΈ

Since sinΞΈcos^2ΞΈ is equal to (1/2)sinΞΈ(2cos^2ΞΈ), which is equal to (1/2)(2sinΞΈcosΞΈcosΞΈ) by using the double angle identity, we can rewrite sinΞΈcos^2ΞΈ as:

(1/2)sin2ΞΈcosΞΈ

Using the identity sin2ΞΈ = 2sinΞΈcosΞΈ, we get:

(1/2)(2sinΞΈcosΞΈ)cosΞΈ

Simplifying further:

(1/2)(2sinΞΈcos^2ΞΈ)

Now, using the identity cos^2ΞΈ = 1 βˆ’ sin^2ΞΈ, we can rewrite cos^2ΞΈ as:

(1/2)(2sinΞΈ(1 βˆ’ sin^2ΞΈ))

Expanding the multiplication gives:

sinΞΈ βˆ’ sin^3ΞΈ

This is equal to the right side of the equation, thus proving the identity:

(1 βˆ’ cos^2ΞΈ)sinΞΈcos^2ΞΈ = 1