Korzystamy ze wzorów redukcyjnych:

$\text{1)}\sin \left({90}^{\circ }-\alpha \right)=\cos \alpha$  

$\text{2)}\cos \left({90}^{\circ }-\alpha \right)=\sin \alpha$  

$\text{3)}\text{tg}\left({90}^{\circ }-\alpha \right)=\text{ctg}\alpha$  

$\text{4)}\text{ctg}\left({90}^{\circ }-\alpha \right)=\text{tg}\alpha$  

gdzie $\alpha \in \left(0^{\circ },{90}^{\circ }\right)$  

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$\text{a)}$  

${\cos }^2{15}^{\circ }-{\cos }^2{30}^{\circ }+{\cos }^2{45}^{\circ }-{\cos }^2{60}^{\circ }+{\cos }^2{75}^{\circ }={\cos }^2{15}^2-{\cos }^2{30}^{\circ }+{\cos }^2{45}^{\circ }-{\left(\cos \left({90}^{\circ }-{30}^{\circ }\right)\right)}^2+{\left(\cos \left({90}^2-{15}^{\circ }\right)\right)}^2=$  

$={\cos }^2{15}^{\circ }-{\cos }^2{30}^{\circ }+{\cos }^2{45}^{\circ }-{\sin }^2{30}^{\circ }+{\sin }^2{15}^{\circ }=\left({\sin }^2{15}^{\circ }+{\cos }^2{15}^{\circ }\right)-\left({\sin }^2{30}^{\circ }+{\cos }^2{30}^{\circ }\right)+{\cos }^2{45}^{\circ }=$

$=1-1+{\left(\frac{\sqrt{2}}{2}\right)}^2=\frac{2}{4}=\frac{1}{2}$  

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$\text{b)}$  

${\sin }^2{10}^{\circ }+{\sin }^2{20}^{\circ }+{\sin }^2{30}^{\circ }+\dots +{\sin }^2{80}^{\circ }=$  

$=\left({\sin }^2{10}^{\circ }+{\sin }^2{80}^{\circ }\right)+\left({\sin }^2{20}^{\circ }+{\sin }^2{70}^{\circ }\right)+\left({\sin }^2{30}^{\circ }+{\sin }^2{60}^{\circ }\right)+\left({\sin }^2{40}^{\circ }+{\sin }^2{50}^{\circ }\right)=$  

$=\left({\sin }^2{10}^{\circ }+{\left(\sin \left({90}^{\circ }-{10}^{\circ }\right)\right)}^2\right)+\left({\sin }^2{20}^{\circ }+{\left(\sin \left({90}^{\circ }-{20}^{\circ }\right)\right)}^2\right)+\left({\sin }^2{30}^{\circ }+{\left(\sin \left({90}^{\circ }-{30}^{\circ }\right)\right)}^2\right)+\left({\sin }^2{40}^{\circ }+{\left(\sin \left({90}^{\circ }-{40}^{\circ }\right)\right)}^2\right)=$  

$=\left({\sin }^2{10}^{\circ }+{\cos }^2{10}^{\circ }\right)+\left({\sin }^2{20}^{\circ }+{\cos }^2{20}^{\circ }\right)+\left({\sin }^2{30}^{\circ }+{\cos }^2{30}^{\circ }\right)+\left({\sin }^2{40}^{\circ }+{\cos }^2{40}^{\circ }\right)=1+1+1+1=4$  

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$\text{c)}$  

$\mathrm{tg}{35}^{\circ }\cdot \mathrm{tg}{40}^{\circ }\cdot \mathrm{tg}{45}^{\circ }\cdot \mathrm{tg}{50}^{\circ }\cdot \mathrm{tg}{55}^{\circ }\cdot \mathrm{tg}{60}^{\circ }=$  

$=\mathrm{tg}{35}^{\circ }\cdot \mathrm{tg}{40}^{\circ }\cdot \mathrm{tg}{45}^{\circ }\cdot \mathrm{tg}\left({90}^{\circ }-{40}^{\circ }\right)\cdot \mathrm{tg}\left({90}^{\circ }-{35}^{\circ }\right)\cdot \mathrm{tg}{60}^{\circ }=$  

$=\underline{\mathrm{tg}{35}^{\circ }}\cdot \underline{\underline{\mathrm{tg}{40}^{\circ }}}\cdot \mathrm{tg}{45}^{\circ }\cdot \underline{\underline{\mathrm{ctg}{40}^{\circ }}}\cdot \underline{\mathrm{ctg}{35}^{\circ }}\cdot \mathrm{tg}{60}^{\circ }=1\cdot 1\cdot \mathrm{tg}{45}^{\circ }\cdot \mathrm{tg}{60}^{\circ }=$  

$=1\cdot 1\cdot 1\cdot \sqrt{3}=\sqrt{3}$