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Cum calculez tg1°×tg2°×tg3°×....tg89°=1?
M-am gandit sa scriu asa: tg1°×tg1×2°×....×tg1×89° si am ajuns undeva unde m.am blocat si anume
tg1°(2×3×4×....×89)=1


Răspuns :

[tex]\text{tg}(90-x^{\circ}) = \text{ctg}(x) \\ \\ \text{tg}(1^{\circ}) \cdot \text{tg}(2^{\circ}) \cdot \text{tg}(3^{\circ}) \cdot ... \cdot \text{tg}(89^\circ) = [/tex]

[tex] = \text{tg}(1^{\circ}) \cdot \text{tg}(2^{\circ}) \cdot \text{tg}(3^{\circ}) \cdot... \cdot \text{tg}(44^{\circ})\cdot \text{tg}(45^{\circ}) \cdot \text{tg}(90^\circ - 44^{\circ})\cdot \\ \cdot \text{tg}(90^{\circ} - 43^{\circ}) \cdot ... \cdot \text{tg}(90^{\circ} - 3^{\circ} ) \cdot \text{tg}(90^{\circ} - 2^{\circ} ) \cdot \text{tg}(90^{\circ} - 1) = [/tex]

[tex]= \text{tg}(1^{\circ}) \cdot \text{tg}(2^{\circ}) \cdot \text{tg}(3^{\circ}) \cdot... \cdot \text{tg}(44^{\circ})\cdot \text{tg}(45^{\circ}) \cdot \text{ctg}(44^{\circ}) \cdot \\ \cdot \text{ctg}(43^{\circ}) \cdot ... \cdot \text{ctg}(3^{\circ}) \cdot \text{ctg}(2^{\circ}) \cdot \text{ctg}(1^{\circ}) =[/tex]

[tex]= \text{tg}(1^{\circ}) \cdot \text{tg}(2^{\circ}) \cdot \text{tg}(3^{\circ}) \cdot... \cdot \text{tg}(44^{\circ})\cdot \text{tg}(45^{\circ}) \cdot \dfrac{1}{\text{tg}(44^{\circ})} \cdot \\ \cdot \dfrac{1}{\text{tg}(43^{\circ})} \cdot ... \cdot \dfrac{1}{\text{tg}(3^{\circ})} \cdot \dfrac{1}{\text{tg}(2^{\circ})} \cdot \dfrac{1}{\text{tg}(1^{\circ})} = \\ \\ = 1\cdot 1\cdot 1\cdot ... \cdot \text{tg}(45^{\circ}) = 1[/tex]