Proving $\lim_{(x,y)\to(\pi,\pi/2)} \frac{\cos y-\sin 2y}{\cos x\cos y}=1$ using $\epsilon$-$\delta$ method

I'm trying to prove the limit$$\lim_{(x,y)\to(\pi,\pi/2)} \frac{\cos y-\sin 2y}{\cos x\cos y}.$$

Since$$\frac{\cos y-\sin 2y}{\cos x\cos y} = \frac{\cos y - 2\sin y\cos y}{\cos x\cos y} = \frac{\cos y(1 - 2\sin y)}{\cos x\cos y} = \frac{1 - 2\sin y}{\cos x},$$the limit is$$\lim_{(x,y)\to(\pi,\pi/2)} \frac{1 - 2\sin y}{\cos x}.$$

Along $x = \pi$,$$\lim_{(x,y)\to(\pi,\pi/2)} \frac{1 - 2\sin y}{\cos x} = \lim_{(x,y)\to(\pi,\pi/2)} 2\sin y - 1 = 1.$$

Along $y=\pi/2$,$$\lim_{(x,y)\to(\pi,\pi/2)} \frac{1 - 2\sin y}{\cos x} = \lim_{(x,y)\to(\pi,\pi/2)} - \frac{1}{\cos x} = 1.$$

Assume that the limit is equal to $1$. For all $\epsilon>0$, we have to find $\delta>0$ s.t.$$0 < \sqrt{\left(x-\pi\right)^2 + \left(y-\frac{\pi}{2}\right)^2} < \delta \implies \left|\frac{1 - 2\sin y}{\cos x} - 1\right| < \epsilon.$$

Now,$$0 < \sqrt{\left(x-\pi\right)^2 + \left(y-\frac{\pi}{2}\right)^2} < \delta \implies 0 < \left|x-\pi\right| < \delta, \quad 0 < \left|y-\frac{\pi}{2}\right| < \delta$$and$$\left|\frac{1 - 2\sin y}{\cos x} - 1\right| = \left|\frac{1 - 2\sin y-\cos x}{\cos x}\right| = \frac{\left|1 - 2\sin y-\cos x\right|}{\left|\cos x\right|}.$$

I tried this way:$$\begin{aligned}\left|1 - 2\sin y-\cos x\right| &= \left|1 - 2\cos \left(\frac{\pi}{2}-y\right) - \sin \left(\frac{\pi}{2}\right)\right| \\&\leq \left|1-\cos \left(\frac{\pi}{2}-y\right)\right| + \left|\cos \left(\frac{\pi}{2}-y\right) + \sin \left(\frac{\pi}{2}-y\right)\right| \\&\leq \left|1-\cos \left(\frac{\pi}{2}-y\right)\right| + 2 \\&\leq 2\left|\sin^2 \left(\frac{\pi}{4}-\frac{y}{2}\right)\right| + 2 \\&\leq \frac{1}{2}\left|y-\frac{\pi}{2}\right| + 2 \\&< \frac{1}{2}\delta^2 + 2 \end{aligned}$$Therefore,$$\left|1 - 2\sin y-\cos x\right| < \frac{1}{2}\delta^2 + 2$$and$$\begin{aligned}\left|\cos x\right| &= \left|\cos \left(\pi-x\right)\right| \\&= \left|1-2\sin^2 \left(\frac{\pi}{2}-\frac{x}{2}\right)\right| \leq 1 + 2\left|\frac{\pi}{2}-\frac{x}{2}\right|^2 \\&= 1 + \frac{1}{2}\left|x-\pi\right|^2 \\&< 1 + \frac{1}{2}\delta^2\end{aligned}$$

Therefore,$$\left|\cos x\right| < 1+\frac{1}{2}\delta^2 \implies \frac{1}{\left|\cos x\right|} > \frac{1}{1+\frac{\delta^2}{2}} = \frac{2}{2+\delta^2}$$But,$$\frac{2}{2+\delta^2} < \frac{1}{\left|\cos x\right|}, \qquad \left|1 - 2\sin y-\cos x\right| < \frac{1}{2}\delta^2 + 2$$so$$\frac{\left|1 - 2\sin y-\cos x\right|}{\left|\cos x\right|} < ???$$

How should I go about it?