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Question Number 57345 by behi83417@gmail.com last updated on 02/Apr/19
1)if: sinx+tgx=1,then: sin^4 x+tg^4 x=? 2)if: sinx+tgx=2,then: sin4x+tg4x=? 3.if: sinx+tgx=3,then: ((sin4x)/(sin^4 x))+((tg4x)/(tg^4 x))=?
$$\left.\mathrm{1}\right)\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{1},\boldsymbol{\mathrm{then}}:\:\:\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{tg}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}=? \\ $$$$\left.\mathrm{2}\right)\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{2},\boldsymbol{\mathrm{then}}:\:\:\boldsymbol{\mathrm{sin}}\mathrm{4}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{tg}}\mathrm{4}\boldsymbol{\mathrm{x}}=? \\ $$$$\mathrm{3}.\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{3},\boldsymbol{\mathrm{then}}:\:\:\frac{\boldsymbol{\mathrm{sin}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}}+\frac{\boldsymbol{\mathrm{tg}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{tg}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}}=? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19
sinx+((sinx)/(cosx))=a ((2t)/(1+t^2 ))+((2t)/(1+t^2 ))×((1+t^2 )/(1−t^2 ))=a ((2t)/(1+t^2 ))+((2t)/(1−t^2 ))=a ((2t−2t^3 +2t+2t^3 )/(1−t^4 ))=a ((4t)/(1−t^4 ))=a a−at^4 −4t=0 at^4 +4t−a=0 t=tan((x/2)) when a=1 t=0.249 when a=2 t=0.475 when a=3 t=0.631 rest of the problem can be solved by putting ((2t)/(1+t^2 )) for sinx and ((2t)/(1−t^2 )) for tanx then put value of t
$${sinx}+\frac{{sinx}}{{cosx}}={a} \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }={a} \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }={a} \\ $$$$\frac{\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}+\mathrm{2}{t}^{\mathrm{3}} }{\mathrm{1}−{t}^{\mathrm{4}} }={a} \\ $$$$\frac{\mathrm{4}{t}}{\mathrm{1}−{t}^{\mathrm{4}} }={a} \\ $$$${a}−{at}^{\mathrm{4}} −\mathrm{4}{t}=\mathrm{0} \\ $$$${at}^{\mathrm{4}} +\mathrm{4}{t}−{a}=\mathrm{0} \\ $$$${t}={tan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$${when}\:{a}=\mathrm{1}\:\:{t}=\mathrm{0}.\mathrm{249} \\ $$$${when}\:{a}=\mathrm{2}\:\:{t}=\mathrm{0}.\mathrm{475} \\ $$$${when}\:{a}=\mathrm{3}\:\:{t}=\mathrm{0}.\mathrm{631} \\ $$$${rest}\:{of}\:{the}\:{problem}\:{can}\:{be}\:{solved}\:{by} \\ $$$${putting}\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{for}\:{sinx}\:{and}\:\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:{for}\:{tanx} \\ $$$${then}\:{put}\:{value}\:{of}\:{t} \\ $$

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