The value of ${\left(\frac{2{\cos }^3\theta -\cos \theta }{\sin \theta -2{\sin }^3\theta }\right)}^2+1,\theta \ne 45°\text{is:}$

The correct option is cosec²θ.

To find the value of the given expression, let us simplify the numerator and denominator of the fraction separately.

The given expression is:

${\left(\frac{2{\cos }^3\theta -\cos \theta }{\sin \theta -2{\sin }^3\theta }\right)}^2+1$

Let us factor out $\cos \theta$ from the numerator and $\sin \theta$ from the denominator:

Numerator:
$2{\cos }^3\theta -\cos \theta =\cos \theta (2{\cos }^2\theta -1)$

Denominator:
$\sin \theta -2{\sin }^3\theta =\sin \theta (1-2{\sin }^2\theta )$

Recall the trigonometric double-angle identities for cosine:
$\cos \left(2\theta \right)=2{\cos }^2\theta -1=1-2{\sin }^2\theta$

Substituting these identities back into the fraction, we get:

$\frac{2{\cos }^3\theta -\cos \theta }{\sin \theta -2{\sin }^3\theta }=\frac{\cos \theta \cdot \cos \left(2\theta \right)}{\sin \theta \cdot \cos \left(2\theta \right)}$

Since we are given that $\theta \ne 45°$, we have $2\theta \ne 90°$, which means $\cos \left(2\theta \right)\ne 0$. We can cancel $\cos \left(2\theta \right)$ from the numerator and denominator:

$\frac{\cos \theta }{\sin \theta }=\cot \theta$

Now, substituting this result back into the main expression:

${\left(\cot \theta \right)}^2+1={\cot }^2\theta +1$

Using the fundamental trigonometric identity $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$, we obtain:

${\cot }^2\theta +1={\mathrm{cosec}}^2\theta$

Thus, the final simplified value of the expression is ${\mathrm{cosec}}^2\theta$.