Prove $|\cos^2(z)| + |\sin^2(z)| > 1$ for complex numbers $z$ with nonzero imaginary part 
Prove $$|\cos^2(z)| + |\sin^2(z)| > 1$$ for $\operatorname{Im}(z) \ne 0$

I know from using the triangle inequality, $|x+y| \leq |x| + |y|$, that $|\cos^2(z)| + |\sin^2(z)| \geq 1$ but I don't know how to carry on to show that it is strictly greater than.
 A: Hint: Recall from geometry that$|z_0-z_2|=|z_0-z_1|+|z_1-z_2|$ means that $z_1$ is on the line between $z_0$ and $z_2$.
We know that $\cos^2(z)=1-\sin^2(z)$, so
$$
\begin{align}
|\sin^2(z)|+|\cos^2(z)|
&=|\sin^2(z)-0|+|1-\sin^2(z)|\\
&=1
\end{align}
$$
means that $\sin^2(z)$ is on the line between $0$ and $1$.

Hint 2: Note that
$$
\begin{align}
\sin(x+iy)&=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\\
|\sin(x+iy)|^2&=\sin^2(x)\cosh^2(y)+\cos^2(x)\sinh^2(y)
\end{align}
$$
and
$$
\begin{align}
\cos(x+iy)&=\cos(x)\cosh(y)-i\sin(x)\sinh(y)\\
|\cos(x+iy)|^2&=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)
\end{align}
$$
Then recall that $\sin^2(x)+\cos^2(x)=1$ and $\cosh^2(y)+\sinh^2(y)=\cosh(2y)$.
A: for $z = x + iy$ with $y< 0$(see the comment) this can be always arranged because either $z$ or its conjugate $\bar z$ has negative imaginary part. we have $\cos(z) + i\sin(z) = e^{iz} = e^{-y}e^{ix}$ now take the absolute value of both sides and squaring it gives $$|\cos z|^2 + |\sin z|^2 = e^{-y} > 1.$$
edit: the above answer is wrong. see the comment by robjohn and my response.
A: Since
$$
\sin(x+iy)=\sin x \cosh y +i\cos x \sinh y \\
\cos(x+iy)=\cos x \cosh y -i\sin x \sinh y,
$$
you have
$$
\left|\sin(x+iy)\right|^2\ge \sin^2 x \cosh^2 y \\
\left|\cos(x+iy)\right|^2\ge \cos^2 x \cosh^2 y,
$$
and
$$
\left|\sin(x+iy)\right|^2 + \left|\cos(x+iy)\right|^2\ge (\cos^2 x + \sin^2 x)\cosh^2 y=\cosh^2 y.
$$
You can conclude that
$$
\left|\sin z\right|^2 + \left|\cos z\right|^2\ge \cosh^2 \left({\text{Im }}z\right).
$$
Since $\cosh x > 1$ for all $x\neq 0$, the desired result follows.
