$a)$  

$Teza:\left(1+\cos x\right)\left(1-\cos x\right)={\sin }^{2}x,x\in \mathbb{R}$  

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

$L=P$  

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$b)$  

$Teza:\cos x{\sin }^{2}x+{\cos }^{3}x=\cos x,x\in \mathbb{R}$  

$L=\cos x{\sin }^{2}x+{\cos }^{3}x=\cos x\left({\sin }^{2}x+{\cos }^{2}x\right)=\cos x\cdot 1=\cos x$  

$L=P$  

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$c)$  

$Teza:1-2\sin x\cos x={\left(\sin x-\cos x\right)}^2,x\in \mathbb{R}$  

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

$P=L$  

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$d)$  

$Teza:{\cos }^{2}x+{\mathrm{tg}}^{2}x{\cos }^{2}x=1,x\in \mathbb{R}\backslash \left\{\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}\right\}$  

$L={\cos }^{2}x+{\mathrm{tg}}^{2}x\cos x^2={\cos }^{2}x+\frac{{\sin }^{2}x}{{\cos }^{2}x}\cdot {\cos }^{2}x={\cos }^{2}x+{\sin }^{2}x=1$   

$L=P$    

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$e)$  

$Teza:\cos x+{\mathrm{tg}}^{2}x\cos x=\frac{1}{\cos x},x\in \mathbb{R}\backslash \left\{\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}\right\}$  

$L=\cos x+{\mathrm{tg}}^{2}x\cos x=\cos x+\frac{{\sin }^{2}x}{{\cos }^{2}x}\cdot \cos x=$  

$=\cos x+\frac{{\sin }^{2}x}{\cos x}=\frac{{\cos }^{2}x+{\sin }^{2}x}{\cos x}=\frac{1}{\cos x}$  

$L=P$  

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$f)$    

$Teza:2{\sin }^{2}x-1=1-2{\cos }^{2}x,x\in \mathbb{R}$  

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

$L=P$  

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$g)$  

$Teza:\frac{1}{{\cos }^{2}x}\left(1-{\sin }^{2}x\right)=1,x\in \mathbb{R}\backslash \left\{\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}\right\}$  

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

$L=P$  

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$h)$  

$Teza:\frac{1}{\sin x\cos x}-\frac{\cos x}{\sin x}=\text{tg}x$  

Dziedzina:

$\sin x\ne 0\wedge \cos x\ne 0$  

skąd mamy 

$x\in \mathbb{R}\backslash \left\{k\cdot \frac{\pi }{2},k\in \mathbb{Z}\right\}$  

 

$L=\frac{1}{\sin x\cos x}-\frac{\cos x}{\sin x}=\frac{1}{\sin x\cos x}-\frac{{\cos }^{2}x}{\sin x\cos x}=$  

$=\frac{1-{\cos }^{2}x}{\sin x\cos x}=\frac{{\sin }^{2}x}{\sin x\cos x}=\frac{\sin x}{\cos x}=\mathrm{tg}x$  

$L=P$    

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$i)$  

$Teza:\frac{\mathrm{tg}x+\cos x}{\cos x\sin x}=\frac{1}{\sin x}+\frac{1}{{\cos }^{2}x}$  

Dziedzina:

$\sin x\ne 0\wedge \cos x\ne 0$  

skąd mamy 

$x\in \mathbb{R}\backslash \left\{k\cdot \frac{\pi }{2},k\in \mathbb{Z}\right\}$  

 

$L=\frac{\mathrm{tg}x+\cos x}{\cos x\sin x}=\frac{\frac{\sin x}{\cos x}+\cos x}{\sin x\cos x}=\frac{\frac{\sin x}{\cos x}+\frac{{\cos }^{2}x}{\cos x}}{\sin x\cos x}=$  

$=\frac{\sin x+{\cos }^{2}x}{\cos x}\cdot \frac{1}{\sin x\cos x}=\frac{\sin x+{\cos }^{2}x}{\sin x{\cos }^{2}x}=\frac{1}{\sin x}+\frac{1}{{\cos }^{2}x}$  

$L=P$  

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$j)$  

$Teza:\frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x}=\frac{2}{\sin x}$  

Dziedzina:

$\sin x\ne 0\wedge \cos x\ne -1$  

skąd mamy 

$x\in \mathbb{R}\backslash \left\{k\pi ,k\in \mathbb{Z}\right\}$  

 

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

$=\frac{1+2\cos x+{\cos }^{2}x+{\sin }^{2}x}{\sin x\left(1+\cos x\right)}=\frac{2+2\cos x}{\sin x\left(1+\cos x\right)}=\frac{2\left(1+\cos x\right)}{\sin x\left(1+\cos x\right)}=\frac{2}{\sin x}$  

$L=P$