Rozpiszmy

$L=\cos \left(\pi -x\right)\cos \left(\pi +x\right)-\sin \left(2\pi -x\right)\sin \left(2\pi +x\right)=\left(-\cos x\right)\left(\cos x\right)-\left(-\sin x\right)\cdot \sin x={\cos }^{2}x+{\sin }^{2}x=1$

$P=\text{tg}x\cdot \text{tg}\left(\frac{\pi }{2}-x\right)=\frac{\sin x}{\cos x}\cdot \frac{\sin \left(\frac{\pi }{2}-x\right)}{\cos \left(\frac{\pi }{2}-x\right)}=\frac{\sin x}{\cos x}\cdot \frac{\cos x}{\sin x}=1$

$L=P$  

c.b.d.u.

Założenia:

$\sin x\ne 0\wedge \cos x\ne 0$  

$x\ne k\pi \wedge x\ne \frac{\pi }{2}+k\pi ,k\in \mathbb{Z}$