Norm of oblique projector and angle between subspaces Take $V$ and $W$ closed subspaces of $H$ a Hilbert space with $V\oplus W=H$ (we'll assume this holds in the sequel, it may not be required everywhere but in the context of interest, it is always verified). 
Define the angle between two subspaces as
$$ \cos\theta_{VW} \quad :=\quad \inf_{v\in V, \|v\|=1} \,\, \|P_{W}v\| $$
with $P_W$ the orthogonal projection onto $W$. Let also $P_{VW}$ denote the oblique projection onto $V$ along $W$ (i.e., $P_{WV}(w+v)=w$ for $w\in W$ and $v\in V$).
There is a first result that I think I understand and which states that $\cos \theta_{VW^\perp}=\cos\theta_{WV^\perp}$. My proof uses that: $\|P_{W^\perp}v\|=\|P_{W^\perp}P_Vv\|$ then that for $\|v\|=1$, $\|P_{W^\perp}P_Vv\|^2+\|P_{W}P_Vv\|^2=1$ which then gives
$$ \cos \theta_{VW^\perp}\quad =\quad \inf_{v\in V,\|v\|=1}\,\,\sqrt{1-\|P_{W}P_Vv\|^2} \quad=\quad \sqrt{1-\|P_WP_V\|_{op}^2} $$
and then we can just use the fact that an orthogonal projector is self-adjoint to get the result.
There is a second result that I do not quite understand and which states that
$$ \|P_{VW}\|\quad=\quad\|I-P_{WV}\| \quad = \quad \sec \theta_{VW^\perp}.  $$
The first equality is trivial but the second troubles me. The first thing that seems strange is that given the previous result we have
$$ \|P_{VW}\|_{op}\quad=\quad\|I-P_{WV}\|_{op} \quad=\quad \sec \theta_{VW^{\perp}}\quad=\quad\sec\theta_{WV^\perp}\quad=\quad\|P_{WV}\|_{op}.$$ 
Second, when trying to prove the result, I can quite easily show that $\|P_{VW}\|_{op}\ge \sec(\theta_{VW^\perp})$ using that for any $w\in W$ and for $\|v\|=1$,
$$ 1 = \|v\| = \|P_{VW}(v-w)\| \le \|P_{VW}\|_{op}\|v-w\| $$
taking $w=P_{W}v$ we get 
$$ {1\over \|P_{VW}\|}_{op} \le \|P_{W^\perp}v\|\quad\text{for any $v\in V$ with $\|v\|=1$} $$
so that $\|P_{VW}\|_{op}^{-1} \le \inf_{v\in V,\|v\|=1}\|P_{W^\perp}v\|=\cos\theta_{VW^{\perp}}$ but I haven't managed to prove the other way..
Any thoughts on this? Thanks a lot in advance.
Ps: this is related to the reading of two articles: mainly "Hilbert Space Idempotents and Involutions" (Bukholtz) article here in which there seems to be a proof of the second result but I haven't managed to follow the steps completely.
There is also "Beyond consistent reconstructions: optimality (...)" (Adcock, Hansen) article on arXiv here
 A: Ok, after spending quite a while on this, I've actually found the answer and post it here hoping it might be useful for someone.

1. $P_{UV} x \quad=\quad P_{UV}P_{V^\perp}x, \,\, \forall x\in H $ i.e., $P_{UV}$ is not modified when restricted to $V^\perp$.
Proof: for $x\in H$, $x=P_V x + (\mathbf{I}-P_V)x$ and by definition of $P_{UV}$, $P_{UV}P_V x=0$ so that $P_{UV}x=P_{UV}(\mathbf{I}-P_V)x$ but $(\mathbf{I}-P_V)=P_{V^\perp}$. $\qquad\blacksquare$

2. $\|P_{UV}\|_{op}\le\left(1-\|P_UP_V\|_{op}^2\right)^{-1/2}$ .
Proof: as we've just seen in 1. we can focus on $x\in V^\perp$. But then
$$ P_{UV} x = \underbrace{x}_{\in V^\perp} + \underbrace{(P_{UV}x-x)}_{\in V}$$
so $\|{P_{UV}x}\|^2=\|{x}\|^2+\|{P_{UV}x-x}\|^2$ ($\star$). But $(P_{UV}x-x)\in V$ so it is equal to $P_V(P_{UV}x-x)=P_VP_{UV}x$ since $x\in V^\perp$. But by definition of $P_{UV}$ we also have $P_{UV}=P_UP_{UV}$ so that $\|P_{UV}x-x\|=\|P_VP_UP_{UV}x\|$. Dividing $(\star)$ by $\|P_UVx\|^2$ and rearranging yields
$$ {\|P_{UV}x\|\over\|x\|} \quad=\quad \left(1-\|P_UP_VP_{UV}x\|^2\right)^{-1/2} $$
Taking the sup over $x\in V^\perp,\|x\|=1$ we get
$$ \|P_{UV}\|_{op} \quad=\quad \sup_{x\in V^\perp,\|x\|=1}\left(1-\|P_UP_VP_{UV}x\|^2\right)^{-1/2} \quad\le\quad \left(1-\|P_UP_V\|_{op}^2\right)^{-1/2}\qquad\blacksquare$$

3. $\cos(\theta_{UV^\perp}) = (1-\|P_UP_V\|_{op}^2)^{1/2}$.
Proof: By definition, $\cos(\theta_{UV^\perp})\doteq \inf_{u\in U,\|u\|=1} \|P_{V^\perp}u\|=\inf_{u\in U,\|u\|=1} \|P_{V^\perp}P_Uu\|$. For any $u\in U$ with $\|u\|=1$ we also have
$$ \|P_{V^\perp P_U u}\|^2+\|P_{V P_U u}\|^2 \quad = \quad 1  $$
so that 
$$ \cos(\theta_{UV^\perp}) \quad=\quad \inf_{u\in U,\|u\|=1} \sqrt{1-\|P_VP_U u\|^2} \quad = \quad \sqrt{1-\|P_VP_U\|^2_{op}}$$
Note that the order of $P_VP_U$ or $P_UP_V$ does not matter ($\|A\|_{op}=\|A^*\|_{op}$ + self-adjoint operators) so we also have $\cos(\theta_{UV^\perp})=\cos(\theta_{VU^\perp})$. $\qquad\blacksquare$

4. $\|P_{UV}\|_{op}\ge \sec(\theta_{UV^\perp})$
Proof: Take $u\in U,\|u\|=1$ and any $v\in V$
$$ 1 = \|u\| = \|P_{UV}(u-v)\| \quad \le\quad \|P_{UV}\|_{op}\|u-v\|  $$
take $v=P_V u$, rearrange, 
$$ {1\over \|P_{UV}\|} \quad\le\quad \|P_{V^\perp}u\|, \qquad \forall u\in U, \|u\|=1 $$ so in particular we can take the inf and have
$${1\over \|P_{UV}\|} \quad\le\quad \inf_{u\in U,\|u\|=1}\|P_{V^\perp}u\| \quad=\quad \cos(\theta_{UV^\perp}).\qquad\blacksquare $$

5. Conclusion
By (2) and (3),
$$ \|P_{UV}\|_{op} \le (1-\|P_UP_V\|^2_{op})^{-1/2} = \sec(\theta_{UV^\perp}). $$
But by (4) $\|P_{UV}\|_{op}\ge\sec(\theta_{UV^\perp})$. So finally we get
$$ \|P_{UV}\|_{op} = \sec(\theta_{UV^\perp}) =\sec(\theta_{VU^\perp}) = \|P_{VU}\|_{op} $$
Remark: this obviously implies that the inequality in (2) is in fact an equality.

The conclusion can actually be related to an identity I was unaware of that states that for any nontrivial projector, one has $\|P\|_{op}=\|\mathbf{I}-P\|_{op}$.
