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Question Number 164788 by cortano1 last updated on 22/Jan/22

Answered by leonhard77 last updated on 22/Jan/22

⇒tan 30° = ((tan 28°+tan 2°)/(1−tan 28.tan 2°°)) ⇒(1/( (√3))) = ((tan 28°+tan 2°)/(1−tan 28°.tan 2°)) 1−tan 28°.tan 2°=(√3) (tan 28°+tan 2°) tan 28°=((tan 26°+tan 2°)/(1−tan 26°.tan 2°)) = ((((√(1−x^2 ))/x) + tan 2°)/(1−((√(1−x^2 ))/x) tan 2°)) = (((√(1−x^2 )) +x tan 2°)/(x−(√(1−x^2 )) tan 2°))

$$\:\Rightarrow\mathrm{tan}\:\mathrm{30}°\:=\:\frac{\mathrm{tan}\:\mathrm{28}°+\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{28}.\mathrm{tan}\:\mathrm{2}°°} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{tan}\:\mathrm{28}°+\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{28}°.\mathrm{tan}\:\mathrm{2}°} \\ $$$$\mathrm{1}−\mathrm{tan}\:\mathrm{28}°.\mathrm{tan}\:\mathrm{2}°=\sqrt{\mathrm{3}}\:\left(\mathrm{tan}\:\mathrm{28}°+\mathrm{tan}\:\mathrm{2}°\right) \\ $$$$ \\ $$$$\mathrm{tan}\:\mathrm{28}°=\frac{\mathrm{tan}\:\mathrm{26}°+\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{26}°.\mathrm{tan}\:\mathrm{2}°} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:+\:\mathrm{tan}\:\mathrm{2}°}{\mathrm{1}−\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\mathrm{tan}\:\mathrm{2}°} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{x}\:\mathrm{tan}\:\mathrm{2}°}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\mathrm{tan}\:\mathrm{2}°} \\ $$

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