$$2 + 2\sec^2{\theta}\tan^2{\theta} - \sec^4{\theta} - \tan^4{\theta}=2+\frac{2*\sin^2\theta}{\cos^4\theta}-\frac{1}{\cos^4\theta}-\frac{\sin^4\theta}{\cos^4\theta}=\frac{2\cos^4\theta-(\sin^4\theta-2\sin^2\theta+1)}{\cos^4\theta}=\frac{2\cos^4\theta-(\sin^2\theta-1)^2}{\cos^4\theta}=\frac{2\cos^4\theta-(\cos^2\theta)^2}{\cos^4\theta}=1$$
So the whole exponent has to be 0 which is easy to tackle with(I guess you don't need help with that)
This is because $\left(\frac{x}{5}\right)^{2} + \left(\frac{y}{2}\right)^{2}$ matches the form of $\cos^{2}t + \sin^{2}t$. But before this, let's understand what $\cos^{2}t + \sin^{2}t = 1$ mean.
We know that the center-radius (standard) form of a circle of radius $r$ with its center at the origin has the equation $$x^{2} + y^{2} = r^{2}.$$
Dividing both sides by $r^{2}$,
$$\begin{align*}\frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} &= 1 \\ \left(\frac{x}{r}\right)^{2} + \left(\frac{y}{r}\right)^{2} &= 1.\end{align*}$$
But we know that $\cos t =\frac{x}{r}$ and $\sin t = \frac{y}{r}$ where $t$ is the angle from the positive $x$-axis. By substitution, we get the equation $$\cos^{2}t + \sin^{2}t = 1 \tag{1}.$$
We will now go to ellipses. Circles as special cases of ellipses where both $x$ and $y$ are scaled by a factor of $r$. However, ellipses that are not circles have different scaling factors for $x$ and $y$. If the scaling factor for $x$ and $y$ are $a$ and $b$, the equation will be $$\left(\frac{x}{a}\right)^{2} + \left(\frac{y}{b}\right)^{2} = 1 \tag{2}.$$
As $(1)$ is similar to $(2)$, we can equate the terms to each other.
\begin{align*}\left(\frac{x}{a}\right)^{2} &= \cos^{2}t &\qquad \left(\frac{y}{b}\right)^{2} &= \sin^{2}t \\ \frac{x}{a} &= \cos t &\qquad \frac{y}{b} &= \sin t \\ x &= a\cos t &\qquad y &= b \sin t.\end{align*}
This is why it works. I can't think of anything aside from this.
Best Answer
Use,
$$\sin^2(\theta)+\cos^2(\theta)=1$$
Dividing both sides by $\cos^2 (\theta)$ gives:
$$\tan ^2 (\theta)+1=\sec^2 (\theta)$$
This should be enough to conclude.
$$x+1=y^2$$