Solve for θ 1+2cot(theta)^2+cot(theta)^4=csc(theta)^n

$1 + 2 \cot^{2} (θ) + \cot^{4} (θ) = {\csc (θ)}^{n}$

Replace the $2 \cot^{2} (θ)$ with $2 (\csc^{2} (θ) - 1)$ based on the $\cot^{2} (x) + 1 = \csc^{2} (x)$ identity.

$1 + 2 (\csc^{2} (θ) - 1) + \cot^{4} (θ) = {\csc (θ)}^{n}$

Simplify each term.

Apply the distributive property.

$1 + 2 \csc^{2} (θ) + 2 \cdot - 1 + \cot^{4} (θ) = {\csc (θ)}^{n}$

Multiply $2$ by $- 1$ .

$1 + 2 \csc^{2} (θ) - 2 + \cot^{4} (θ) = {\csc (θ)}^{n}$

Subtract $2$ from $1$ .

$2 \csc^{2} (θ) - 1 + \cot^{4} (θ) = {\csc (θ)}^{n}$

Reorder the polynomial.

$\cot^{4} (θ) + 2 \csc^{2} (θ) - 1 = {\csc (θ)}^{n}$

Subtract ${\csc (θ)}^{n}$ from both sides of the equation.

$\cot^{4} (θ) + 2 \csc^{2} (θ) - 1 - {\csc (θ)}^{n} = 0$

Replace the $2 \csc^{2} (θ)$ with $2 (\cot^{2} (θ) + 1)$ based on the $\cot^{2} (x) + 1 = \csc^{2} (x)$ identity.

$\cot^{4} (θ) + 2 (\cot^{2} (θ) + 1) - 1 - {\csc (θ)}^{n} = 0$

Simplify each term.

Apply the distributive property.

$\cot^{4} (θ) + 2 \cot^{2} (θ) + 2 \cdot 1 - 1 - {\csc (θ)}^{n} = 0$

Multiply $2$ by $1$ .

$\cot^{4} (θ) + 2 \cot^{2} (θ) + 2 - 1 - {\csc (θ)}^{n} = 0$

Subtract $1$ from $2$ .

$\cot^{4} (θ) + 2 \cot^{2} (θ) + 1 - {\csc (θ)}^{n} = 0$

Subtract $1$ from both sides of the equation.

$\cot^{4} (θ) + 2 \cot^{2} (θ) - {\csc (θ)}^{n} = - 1$