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Question Number 79437 by Vishal Sharma last updated on 25/Jan/20
The value of tan 1° tan 2° tan 3°...tan 89°  is
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\mathrm{1}°\:\mathrm{tan}\:\mathrm{2}°\:\mathrm{tan}\:\mathrm{3}°…\mathrm{tan}\:\mathrm{89}° \\ $$$$\mathrm{is} \\ $$
Commented by mr W last updated on 25/Jan/20
=1  since tan (x)×tan (90−x)=1
$$=\mathrm{1} \\ $$$${since}\:\mathrm{tan}\:\left({x}\right)×\mathrm{tan}\:\left(\mathrm{90}−{x}\right)=\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 25/Jan/20
A =tan((π/(180))).tan(((2π)/(180))).....tan(((89π)/(180)))  we have tan(((89π)/(180)))=tan((π/2)−(π/(180)))=(1/(tan((π/(180)))))  tan(((88π)/(180)))=tan((π/2)−((2π)/(180)))=(1/(tan(((2π)/(180))))) ....⇒A =1
$${A}\:={tan}\left(\frac{\pi}{\mathrm{180}}\right).{tan}\left(\frac{\mathrm{2}\pi}{\mathrm{180}}\right)…..{tan}\left(\frac{\mathrm{89}\pi}{\mathrm{180}}\right) \\ $$$${we}\:{have}\:{tan}\left(\frac{\mathrm{89}\pi}{\mathrm{180}}\right)={tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{180}}\right)=\frac{\mathrm{1}}{{tan}\left(\frac{\pi}{\mathrm{180}}\right)} \\ $$$${tan}\left(\frac{\mathrm{88}\pi}{\mathrm{180}}\right)={tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}\pi}{\mathrm{180}}\right)=\frac{\mathrm{1}}{{tan}\left(\frac{\mathrm{2}\pi}{\mathrm{180}}\right)}\:….\Rightarrow{A}\:=\mathrm{1} \\ $$

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