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Question Number 115341 by bemath last updated on 25/Sep/20
If log tan 1°+log tan 2°+log tan 3°+...+log tan 89°=p  then p^2 +3 =
$${If}\:\mathrm{log}\:\mathrm{tan}\:\mathrm{1}°+\mathrm{log}\:\mathrm{tan}\:\mathrm{2}°+\mathrm{log}\:\mathrm{tan}\:\mathrm{3}°+…+\mathrm{log}\:\mathrm{tan}\:\mathrm{89}°={p} \\ $$$${then}\:{p}^{\mathrm{2}} +\mathrm{3}\:=\: \\ $$
Answered by bobhans last updated on 25/Sep/20
⇒log (tan 1°×tan 2°×tan 3°×...×tan 89°)=p  consider tan 89°×tan 1°=1  tan 88°×tan 2°=1 , so on . we get   log (1×1×1×1×...×1) = p  ⇒p = 0 then p^2 +3 = 3
$$\Rightarrow\mathrm{log}\:\left(\mathrm{tan}\:\mathrm{1}°×\mathrm{tan}\:\mathrm{2}°×\mathrm{tan}\:\mathrm{3}°×…×\mathrm{tan}\:\mathrm{89}°\right)={p} \\ $$$${consider}\:\mathrm{tan}\:\mathrm{89}°×\mathrm{tan}\:\mathrm{1}°=\mathrm{1} \\ $$$$\mathrm{tan}\:\mathrm{88}°×\mathrm{tan}\:\mathrm{2}°=\mathrm{1}\:,\:{so}\:{on}\:.\:{we}\:{get}\: \\ $$$$\mathrm{log}\:\left(\mathrm{1}×\mathrm{1}×\mathrm{1}×\mathrm{1}×…×\mathrm{1}\right)\:=\:{p} \\ $$$$\Rightarrow{p}\:=\:\mathrm{0}\:{then}\:{p}^{\mathrm{2}} +\mathrm{3}\:=\:\mathrm{3} \\ $$

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