$(1)$ \[ P-F\cos\alpha-N\sin\alpha=0\\ \therefore P=F\cos\alpha+N\sin\alpha \hspace{10pt}\cdots(1)' \] $(2)$ \[ Q+F\sin\alpha-N\cos\alpha=0\\ \therefore Q=-F\sin\alpha+N\cos\alpha \hspace{10pt}\cdots(2)' \] $(3)$ %=image:/media/2015/01/22/142186061670686900.png: $(4)$ \[ \tan\lambda = \frac{F}{N} \ より \ F=N\cdot \tan\lambda\\ \] 式$(1)'\cdot(2)'$に代入すると， \[\begin{align} P &=N \tan\lambda\cdot \cos\alpha+N\sin\alpha\\ &=N\left( \tan\lambda \cdot \cos\alpha +\sin\alpha \right)\cdots(1)' '\\ \end{align} \] \[\begin{align} Q &=-N \tan\lambda\cdot \sin\alpha+N \cos\alpha\\ &=N\left( \cos\alpha - \tan\lambda \cdot \sin\alpha \right)\cdots(2)' '\\ \end{align}\] 式$(2)' '$より， \[ N=\frac{Q}{\cos\alpha-\tan\lambda\cdot \sin\alpha} \] 式$(1)' '$に代入すると， \[\begin{align} P&=\frac{Q\left(\tan\lambda \cdot \cos\alpha +\sin\alpha \right)}{\cos\alpha - \tan\lambda\cdot \sin\alpha}\\ &=Q\frac{\tan\lambda+\frac{\sin\alpha}{\cos\alpha}}{1-\tan\lambda\cdot\frac{\sin\alpha}{\cos\alpha}}\\ &=Q\frac{\\tan\lambda+\tan\alpha}{1-\tan\lambda\cdot \tan\alpha}\\ &=Q\cdot \tan\left( \alpha + \lambda \right) \end{align}\] \[ \therefore P=Q \tan\left( \alpha + \lambda \right) \]