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Question-14077

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Question Number 14077 by Tinkutara last updated on 27/May/17
Answered by ajfour last updated on 28/May/17
let (π/4)cot θ=α and (π/4)tan θ=β then sin α=cos β=f (say) β=2pπ±cos^(−1) f ; p ∈ Z ....(a) α=mπ+(−1)^m sin^(−1) f ; m ∈ Z ..(b) α+β=(π/4)(cot θ+tan θ) ⇒ α+β=(π/(2sin 2θ)) ......(i) α−β=(π/4)(cot θ−tan θ) ⇒ α−β=(π/(2tan 2θ)) .......(ii) comparing (b)+(a) with (i): (π/(2sin 2θ))=(2p+m)π±cos^(−1) f +(−1)^m sin^(−1) f ...(I) comparing (b)−(a) with (ii): (π/(2tan 2θ))=(m−2p)π+(−1)^m sin^(−1) f ∓cos^(−1) f ...(II) f is to be eliminated , four cases arise from (I) & (II): In eqn (I) let m is ′even′ and let us first use +cos^(−1) f ..then (π/(2sin 2θ))=2nπ+(π/2) ⇒ (1/(sin 2θ))=4n+1 ⇒ sin 2θ=(1/(4n+1)) ⇒ 2θ=kπ+(−1)^k sin^(−1) ((1/(4n+1))) or 𝛉=((k𝛑)/2)+(((−1)^n )/2)sin^(−1) ((1/(4n+1))) In ean (II) let m is ′even′ and let us use first +cos^(−1) f ...then (π/(2tan 2θ))=−2nπ+(π/2) (1/(tan 2θ))=−4n+1 ⇒ tan 2θ=((−1)/(4n−1)) 2θ=kπ+tan^(−1) (((−1)/(4n−1))) or 𝛉=((k𝛑)/2)+(1/2)tan^(−1) (((−1)/(4n−1))) in such a way two more solutions are possible ;(are not in your book, as you indicated).
$${let}\:\frac{\pi}{\mathrm{4}}\mathrm{cot}\:\theta=\alpha\:\:{and}\:\:\frac{\pi}{\mathrm{4}}\mathrm{tan}\:\theta=\beta \\ $$$$\:\:{then}\:\:\mathrm{sin}\:\alpha=\mathrm{cos}\:\beta={f}\:\:\left({say}\right) \\ $$$$\beta=\mathrm{2}{p}\pi\pm\mathrm{cos}^{−\mathrm{1}} {f}\:\:;\:\:{p}\:\in\:{Z}\:\:\:\:\:\:….\left({a}\right) \\ $$$$\alpha={m}\pi+\left(−\mathrm{1}\right)^{{m}} \mathrm{sin}^{−\mathrm{1}} {f}\:\:;\:{m}\:\in\:{Z}\:..\left({b}\right) \\ $$$$\alpha+\beta=\frac{\pi}{\mathrm{4}}\left(\mathrm{cot}\:\theta+\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\:\:\:\alpha+\beta=\frac{\pi}{\mathrm{2sin}\:\mathrm{2}\theta}\:\:\:\:\:\:\:\:……\left({i}\right) \\ $$$$\alpha−\beta=\frac{\pi}{\mathrm{4}}\left(\mathrm{cot}\:\theta−\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\:\:\:\alpha−\beta=\frac{\pi}{\mathrm{2tan}\:\mathrm{2}\theta}\:\:\:\:\:\:\:…….\left({ii}\right)\:\:\:\: \\ $$$${comparing}\:\left({b}\right)+\left({a}\right)\:{with}\:\left({i}\right): \\ $$$$\frac{\pi}{\mathrm{2sin}\:\mathrm{2}\theta}=\left(\mathrm{2}{p}+{m}\right)\pi\pm\mathrm{cos}^{−\mathrm{1}} {f} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(−\mathrm{1}\right)^{{m}} \mathrm{sin}^{−\mathrm{1}} {f}\:\:\:\:\:\:…\left({I}\right) \\ $$$${comparing}\:\left({b}\right)−\left({a}\right)\:{with}\:\left({ii}\right): \\ $$$$\frac{\pi}{\mathrm{2tan}\:\mathrm{2}\theta}=\left({m}−\mathrm{2}{p}\right)\pi+\left(−\mathrm{1}\right)^{{m}} \mathrm{sin}^{−\mathrm{1}} {f} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mp\mathrm{cos}^{−\mathrm{1}} {f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({II}\right) \\ $$$$\:\:\boldsymbol{{f}}\:{is}\:{to}\:{be}\:{eliminated}\:,\:{four}\:{cases} \\ $$$${arise}\:{from}\:\:\left({I}\right)\:\&\:\left({II}\right): \\ $$$$ \\ $$$${In}\:{eqn}\:\left({I}\right)\:{let}\:{m}\:{is}\:'{even}'\:{and} \\ $$$$\:\:{let}\:{us}\:{first}\:{use}\:+\mathrm{cos}^{−\mathrm{1}} {f}\:..{then} \\ $$$$\frac{\pi}{\mathrm{2sin}\:\mathrm{2}\theta}=\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\theta}=\mathrm{4}{n}+\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{2}\theta={k}\pi+\left(−\mathrm{1}\right)^{{k}} \mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{1}}\right) \\ $$$$\:{or}\:\boldsymbol{\theta}=\frac{\boldsymbol{{k}\pi}}{\mathrm{2}}+\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}\boldsymbol{{n}}+\mathrm{1}}\right) \\ $$$${In}\:{ean}\:\left({II}\right)\:{let}\:{m}\:{is}\:'{even}'\:{and} \\ $$$${let}\:{us}\:{use}\:{first}\:+\mathrm{cos}^{−\mathrm{1}} {f}\:…{then} \\ $$$$\frac{\pi}{\mathrm{2tan}\:\mathrm{2}\theta}=−\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\theta}=−\mathrm{4}{n}+\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{tan}\:\mathrm{2}\theta=\frac{−\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}} \\ $$$$\mathrm{2}\theta={k}\pi+\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}\right) \\ $$$${or}\:\:\boldsymbol{\theta}=\frac{\boldsymbol{{k}\pi}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\mathrm{4}\boldsymbol{{n}}−\mathrm{1}}\right)\: \\ $$$${in}\:{such}\:{a}\:{way}\:{two}\:{more} \\ $$$${solutions}\:{are}\:{possible}\:;\left({are}\:{not}\right. \\ $$$$\left.{in}\:{your}\:{book},\:{as}\:{you}\:{indicated}\right). \\ $$
Answered by Tinkutara last updated on 27/Jul/17
cos ((π/4) tan θ) = cos ((π/2) − (π/4) cot θ) (π/4) tan θ = 2nπ ± ((π/2) − (π/4) cot θ) Taking positive sign, (1/4)(tan θ + cot θ) = 2n + (1/2) ((1 + tan^2 θ)/(2 tan θ)) = 4n + 1 = cosec 2θ 2θ = nπ + (−1)^n cosec^(−1) (4n + 1) θ = ((nπ)/2) + (((−1)^n )/2) sin^(−1) ((1/(4n + 1))) Now taking negative sign, (1/4)(tan θ − cot θ) = 2n − (1/2) ((tan^2 θ − 1)/(2 tan θ)) = 4n − 1 = − cot 2θ 2θ = nπ + cot^(−1) [−(4n − 1)] θ = ((nπ)/2) + (1/2) tan^(−1) (((−1)/(4n − 1)))
$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\:\mathrm{tan}\:\theta\right)\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\:−\:\frac{\pi}{\mathrm{4}}\:\mathrm{cot}\:\theta\right) \\ $$$$\frac{\pi}{\mathrm{4}}\:\mathrm{tan}\:\theta\:=\:\mathrm{2}{n}\pi\:\pm\:\left(\frac{\pi}{\mathrm{2}}\:−\:\frac{\pi}{\mathrm{4}}\:\mathrm{cot}\:\theta\right) \\ $$$$\mathrm{Taking}\:\mathrm{positive}\:\mathrm{sign}, \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{tan}\:\theta\:+\:\mathrm{cot}\:\theta\right)\:=\:\mathrm{2}{n}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{2}} \:\theta}{\mathrm{2}\:\mathrm{tan}\:\theta}\:=\:\mathrm{4}{n}\:+\:\mathrm{1}\:=\:\mathrm{cosec}\:\mathrm{2}\theta \\ $$$$\mathrm{2}\theta\:=\:{n}\pi\:+\:\left(−\mathrm{1}\right)^{{n}} \mathrm{cosec}^{−\mathrm{1}} \left(\mathrm{4}{n}\:+\:\mathrm{1}\right) \\ $$$$\theta\:=\:\frac{{n}\pi}{\mathrm{2}}\:+\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}{n}\:+\:\mathrm{1}}\right) \\ $$$$\mathrm{Now}\:\mathrm{taking}\:\mathrm{negative}\:\mathrm{sign}, \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{tan}\:\theta\:−\:\mathrm{cot}\:\theta\right)\:=\:\mathrm{2}{n}\:−\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{tan}^{\mathrm{2}} \:\theta\:−\:\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\theta}\:=\:\mathrm{4}{n}\:−\:\mathrm{1}\:=\:−\:\mathrm{cot}\:\mathrm{2}\theta \\ $$$$\mathrm{2}\theta\:=\:{n}\pi\:+\:\mathrm{cot}^{−\mathrm{1}} \left[−\left(\mathrm{4}{n}\:−\:\mathrm{1}\right)\right] \\ $$$$\theta\:=\:\frac{{n}\pi}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\mathrm{4}{n}\:−\:\mathrm{1}}\right) \\ $$

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