Odd One Out

Taking a look at the remaining choices,

$\large \frac {\sin^6 \theta - \cos^6 \theta}{\sin^2 \theta \cos^4 \theta - \cos^6 \theta}$

Multiplying $\large \sec^6 \theta$ to both the numerator and denominator,

$\large \frac {\tan^6 \theta - 1}{\tan^2 \theta - 1}$

$\large \tan^4 \theta + \tan^2 \theta + 1$

...which is the same another choice: $\large \large \tan^4 \theta + \sec^2 \theta$ .

On the other hand, $\large \large \sec^4 \theta - \tan^2 \theta$ can be converted to:

$\large (\sec^2 \theta)^2 - \tan^2 \theta$

$\large (\tan^2 \theta + 1)^2 - \tan^2 \theta$

$\large \tan^4 \theta + 2\tan^2 \theta + 1 - \tan^2 \theta$

$\large \tan^4 \theta + \tan^2 \theta + 1$

...which results to: $\large \tan^4 \theta + \sec^2 \theta$ .

$\large \sec^4 \theta ( 1 + \frac {\sin^2 2\theta}{4} )$ is different because of the plus sign inside the parentheses. If it was minus sign, it would be the same as $\large \sec^4 \theta - \tan^2 \theta$ .