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Question-22730




Question Number 22730 by selestian last updated on 22/Oct/17
Commented by math solver last updated on 22/Oct/17
−1. we know tan is positive in   1st and 3rd quadrant .  so there are infinite values which  we can take .  say we take 60 degree and 240 degree.
$$−\mathrm{1}.\:{we}\:{know}\:{tan}\:{is}\:{positive}\:{in}\: \\ $$$$\mathrm{1}{st}\:{and}\:\mathrm{3}{rd}\:{quadrant}\:. \\ $$$${so}\:{there}\:{are}\:{infinite}\:{values}\:{which} \\ $$$${we}\:{can}\:{take}\:. \\ $$$${say}\:{we}\:{take}\:\mathrm{60}\:{degree}\:{and}\:\mathrm{240}\:{degree}. \\ $$
Commented by selestian last updated on 22/Oct/17
yes you are right thAnks i got the answer
$${yes}\:{you}\:{are}\:{right}\:{thAnks}\:{i}\:{got}\:{the}\:{answer} \\ $$$$ \\ $$
Answered by ajfour last updated on 22/Oct/17
let the smaller angle be θ_1 .       0 ≤θ_1 ≤ π  Then θ_2 =θ_1 +π  ⇒ tan (θ_2 /2)= tan ((θ_1 /2)+(π/2))  ⇒ tan (θ_2 /2)=−cot (θ_1 /2)  ⇒  tan (θ_1 /2)tan (θ_2 /2) =−1 .
$${let}\:{the}\:{smaller}\:{angle}\:{be}\:\theta_{\mathrm{1}} . \\ $$$$\:\:\:\:\:\mathrm{0}\:\leqslant\theta_{\mathrm{1}} \leqslant\:\pi \\ $$$${Then}\:\theta_{\mathrm{2}} =\theta_{\mathrm{1}} +\pi \\ $$$$\Rightarrow\:\mathrm{tan}\:\frac{\theta_{\mathrm{2}} }{\mathrm{2}}=\:\mathrm{tan}\:\left(\frac{\theta_{\mathrm{1}} }{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{tan}\:\frac{\theta_{\mathrm{2}} }{\mathrm{2}}=−\mathrm{cot}\:\frac{\theta_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\frac{\theta_{\mathrm{1}} }{\mathrm{2}}\mathrm{tan}\:\frac{\theta_{\mathrm{2}} }{\mathrm{2}}\:=−\mathrm{1}\:. \\ $$

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