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[winfast29]

\(\displaystyle{ cos(90- ) \frac{ctg(360- )}{sin(270+ )}=1}\)

[Nakahed90]

\(\displaystyle{ L=\cos (90-\alpha)*\frac{\ctg (360-\alpha)}{\sin (270+\alpha)}=\sin *\frac{-\ctg }{-\cos }=\frac{\sin *\ctg }{\cos }=\frac{\sin *\frac{\cos }{\sin }}{\cos }=1=P}\)