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Question Number 177791 by cortano1 last updated on 09/Oct/22
Prove that 1−cot 23° = (2/(1−cot 22°))
$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\mathrm{cot}\:\mathrm{23}°\:=\:\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cot}\:\mathrm{22}°}\: \\ $$
Answered by som(math1967) last updated on 09/Oct/22
cot45=1 cot(23+22)=1 ⇒((cot22cot23−1)/(cot23+cot22))=1 ⇒cot23cot22=1+cot22+cot23 ⇒−cot23(1−cot22)=1+cot22 ⇒−cot23=((1+cot22)/(1−cot22)) ⇒1−cot23=1+((1+cot22)/(1−cot22)) ∴ 1−cot23=(2/(1−cot22))
$${cot}\mathrm{45}=\mathrm{1} \\ $$$$\:{cot}\left(\mathrm{23}+\mathrm{22}\right)=\mathrm{1} \\ $$$$\Rightarrow\frac{{cot}\mathrm{22}{cot}\mathrm{23}−\mathrm{1}}{{cot}\mathrm{23}+{cot}\mathrm{22}}=\mathrm{1} \\ $$$$\Rightarrow{cot}\mathrm{23}{cot}\mathrm{22}=\mathrm{1}+{cot}\mathrm{22}+{cot}\mathrm{23} \\ $$$$\Rightarrow−{cot}\mathrm{23}\left(\mathrm{1}−{cot}\mathrm{22}\right)=\mathrm{1}+{cot}\mathrm{22}\: \\ $$$$\Rightarrow−{cot}\mathrm{23}=\frac{\mathrm{1}+{cot}\mathrm{22}}{\mathrm{1}−{cot}\mathrm{22}} \\ $$$$\Rightarrow\mathrm{1}−{cot}\mathrm{23}=\mathrm{1}+\frac{\mathrm{1}+{cot}\mathrm{22}}{\mathrm{1}−{cot}\mathrm{22}} \\ $$$$\therefore\:\mathrm{1}−{cot}\mathrm{23}=\frac{\mathrm{2}}{\mathrm{1}−{cot}\mathrm{22}} \\ $$
Answered by blackmamba last updated on 09/Oct/22
tan 22°=((1−tan 23°)/(1+tan 23°)) cot 22°= ((1+tan 23°)/(1−tan 23°)) ⇒1−cot 22°=1−(((1+tan 23°)/(1−tan 23°))) ⇒ 1−cot 22° = ((−2tan 23° )/(1−tan 23°)) ⇒(1−cot 23°)×(1−cot 22°) = (((tan 23°−1)/(tan 23°)))×(((−2tan 23°)/(1−tan 23°))) = 2 (proved)
$$\:\mathrm{tan}\:\mathrm{22}°=\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°}{\mathrm{1}+\mathrm{tan}\:\mathrm{23}°} \\ $$$$\:\mathrm{cot}\:\mathrm{22}°=\:\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°} \\ $$$$\:\Rightarrow\mathrm{1}−\mathrm{cot}\:\mathrm{22}°=\mathrm{1}−\left(\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°}\right) \\ $$$$\Rightarrow\:\mathrm{1}−\mathrm{cot}\:\mathrm{22}°\:=\:\frac{−\mathrm{2tan}\:\mathrm{23}°\:}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{cot}\:\mathrm{23}°\right)×\left(\mathrm{1}−\mathrm{cot}\:\mathrm{22}°\right) \\ $$$$\:=\:\left(\frac{\mathrm{tan}\:\mathrm{23}°−\mathrm{1}}{\mathrm{tan}\:\mathrm{23}°}\right)×\left(\frac{−\mathrm{2tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°}\right) \\ $$$$\:=\:\mathrm{2}\:\left({proved}\right) \\ $$

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