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Question Number 135795 by benjo_mathlover last updated on 16/Mar/21
What the value of (1−cot 23°)(1−cot 22°).
$${What}\:{the}\:{value}\:{of}\:\left(\mathrm{1}−\mathrm{cot}\:\mathrm{23}°\right)\left(\mathrm{1}−\mathrm{cot}\:\mathrm{22}°\right). \\ $$
Answered by MJS_new last updated on 16/Mar/21
(1−cot (((45)/2)−x))(1−cot (((45)/2)+x))=2
$$\left(\mathrm{1}−\mathrm{cot}\:\left(\frac{\mathrm{45}}{\mathrm{2}}−{x}\right)\right)\left(\mathrm{1}−\mathrm{cot}\:\left(\frac{\mathrm{45}}{\mathrm{2}}+{x}\right)\right)=\mathrm{2} \\ $$
Answered by liberty last updated on 16/Mar/21
cot 22° = cot (45°−23°) cot 22°= (1/(tan (45°−23°))) = (1/((1−tan 23°)/(1+tan 23°))) cot 22° = ((1+tan 23°)/(1−tan 23°)) so 1−cot 22°=1−((1+tan 23°)/(1−tan 23°))=−((2tan 23°)/(1−tan 23°)) then (1−(1/(tan 23°)))(−((2tan 23°)/(1−tan 23°))) = ((tan 23°−1)/(tan 23°))×((−2tan 23°)/(1−tan 23°)) = 2
$$\mathrm{cot}\:\mathrm{22}°\:=\:\mathrm{cot}\:\left(\mathrm{45}°−\mathrm{23}°\right) \\ $$$$\:\mathrm{cot}\:\mathrm{22}°=\:\frac{\mathrm{1}}{\mathrm{tan}\:\left(\mathrm{45}°−\mathrm{23}°\right)}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°}{\mathrm{1}+\mathrm{tan}\:\mathrm{23}°}} \\ $$$$\mathrm{cot}\:\mathrm{22}°\:=\:\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°} \\ $$$${so}\:\mathrm{1}−\mathrm{cot}\:\mathrm{22}°=\mathrm{1}−\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°}=−\frac{\mathrm{2tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°} \\ $$$${then}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{23}°}\right)\left(−\frac{\mathrm{2tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°}\right) \\ $$$$=\:\frac{\mathrm{tan}\:\mathrm{23}°−\mathrm{1}}{\mathrm{tan}\:\mathrm{23}°}×\frac{−\mathrm{2tan}\:\mathrm{23}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{23}°}\:=\:\mathrm{2} \\ $$

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