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Question Number 19385 by NEC last updated on 10/Aug/17

tan^2 β=−1 find β.... lets solve for fun

$${tan}^{\mathrm{2}} \beta=−\mathrm{1} \\ $$$$ \\ $$$${find}\:\beta....\:{lets}\:{solve}\:{for}\:{fun} \\ $$

Commented by NEC last updated on 10/Aug/17

please help

$${please}\:{help} \\ $$

Commented by Tinkutara last updated on 10/Aug/17

There is no such β (real or complex).

$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{such}\:\beta\:\left(\mathrm{real}\:\mathrm{or}\:\mathrm{complex}\right). \\ $$

Commented by NEC last updated on 10/Aug/17

the β could also be represented by x or θ or any value. A friend actually gave me the question i′m stocked. this was what i did tan^2 x=−1 tan^2 x + 1=0 sec^2 x=0 (1/(cos^2 x))=0 from that point i became confused..... i know it might lead to a complex number but if its possible i′ll like to see the solution

$${the}\:\beta\:{could}\:{also}\:{be}\:{represented}\:{by} \\ $$$${x}\:\boldsymbol{{or}}\:\theta\:\boldsymbol{{or}}\:\boldsymbol{{any}}\:\boldsymbol{{value}}. \\ $$$$ \\ $$$$\boldsymbol{{A}}\:{friend}\:{actually}\:{gave}\:{me}\:{the} \\ $$$${question}\:{i}'{m}\:{stocked}. \\ $$$$ \\ $$$${this}\:{was}\:{what}\:{i}\:{did} \\ $$$$\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}=−\mathrm{1} \\ $$$$\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}\:+\:\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{became}} \\ $$$$\boldsymbol{\mathrm{confused}}..... \\ $$$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{know}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{might}}\:\boldsymbol{\mathrm{lead}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{complex}} \\ $$$$\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{but}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{i}}'{ll}\:{like} \\ $$$${to}\:{see}\:{the}\:{solution} \\ $$

Commented by Tinkutara last updated on 10/Aug/17

It is impossible. Let z = e^(iθ) = cos θ + i sin θ and z^� = e^(−iθ) = cos θ − i sin θ. z + z^� = 2 cos θ = e^(iθ) + e^(−iθ) (1/(cos θ)) = (2/(e^(iθ) + e^(−iθ) )) = ((2e^(iθ) )/(e^(2iθ) + 1)) and this equal to 0 implies that e^(iθ) = 0 which is impossible.

$$\mathrm{It}\:\mathrm{is}\:\mathrm{impossible}.\:\mathrm{Let}\:{z}\:=\:{e}^{{i}\theta} \:=\:\mathrm{cos}\:\theta\:+ \\ $$$${i}\:\mathrm{sin}\:\theta\:\mathrm{and}\:\bar {{z}}\:=\:{e}^{−{i}\theta} \:=\:\mathrm{cos}\:\theta\:−\:{i}\:\mathrm{sin}\:\theta. \\ $$$${z}\:+\:\bar {{z}}\:=\:\mathrm{2}\:\mathrm{cos}\:\theta\:=\:{e}^{{i}\theta} \:+\:{e}^{−{i}\theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\:=\:\frac{\mathrm{2}}{{e}^{{i}\theta} \:+\:{e}^{−{i}\theta} }\:=\:\frac{\mathrm{2}{e}^{{i}\theta} }{{e}^{\mathrm{2}{i}\theta} \:+\:\mathrm{1}}\:\mathrm{and}\:\mathrm{this} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{0}\:\mathrm{implies}\:\mathrm{that}\:{e}^{{i}\theta} \:=\:\mathrm{0}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{impossible}. \\ $$

Commented by NEC last updated on 10/Aug/17

thanks

$${thanks} \\ $$

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