$a)\sqrt{1+{\mathrm{tg}}^{2}x}=-\frac{1}{\cos x}$  

$L=\sqrt{\frac{{\cos }^{2}x}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}}=\sqrt{\frac{{\sin }^{2}x+{\cos }^{2}x}{{\cos }^{2}x}}=\sqrt{\frac{1}{{\cos }^{2}x}}=\left|\frac{1}{\cos x}\right|=\frac{1}{\left|\cos x\right|}=$ $L=\sqrt{\frac{{\cos }^{2}x}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}}=\sqrt{\frac{{\sin }^{2}x+{\cos }^{2}x}{{\cos }^{2}x}}=\sqrt{\frac{1}{{\cos }^{2}x}}=\left|\frac{1}{\cos x}\right|=\frac{1}{\left|\cos x\right|}=$  

$\text{Dla}x\in \left(\frac{\pi }{2},\pi \right)\cos x<0$  

$=-\frac{1}{\cos x}=P$  

 

$b)\sqrt{1+{\mathrm{ctg}}^{2}x}=\frac{1}{\sin x}$  

$L=\sqrt{\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}}=\sqrt{\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}}=\sqrt{\frac{1}{{\sin }^{2}x}}=\left|\frac{1}{\sin x}\right|=\frac{1}{\left|\sin x\right|}=$  

$\text{Dla}x\in \left(\frac{\pi }{2},\pi \right)\sin x>0$  

$=\frac{1}{\sin x}=P$