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Question Number 39312 by kunal1234523 last updated on 05/Jul/18
prove that (tan 4a+tan 2a)(1−tan^2 3a tan^2 a)=2tan 3a sec^2 a
$${prove}\:{that} \\ $$$$\left({tan}\:\mathrm{4}{a}+{tan}\:\mathrm{2}{a}\right)\left(\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}{a}\:{tan}^{\mathrm{2}} {a}\right)=\mathrm{2}{tan}\:\mathrm{3}{a}\:{sec}^{\mathrm{2}} {a} \\ $$
Answered by kunal1234523 last updated on 05/Jul/18
Commented by kunal1234523 last updated on 05/Jul/18
another way please
$${another}\:{way}\:{please} \\ $$
Commented by abdo mathsup 649 cc last updated on 08/Jul/18
another way is going to hospital of crazy...
$${another}\:{way}\:{is}\:{going}\:{to}\:{hospital}\:{of}\:{crazy}… \\ $$
Answered by MJS last updated on 05/Jul/18
let me write c_n /s_n /t_n for cos nx/sin nx/tan nx (t_4 +t_2 )(1−t_3 ^2 t^2 )=2(t_3 /c^2 ) t_2 +t_4 −t^2 t_3 ^2 t_4 −t^2 t_2 t_3 ^2 =2(t_3 /c^2 ) (s_2 /c_2 )+(s_4 /c_4 )−((s^2 s_3 ^2 s_4 )/(c^2 c_3 ^2 c_4 ))−((s^2 s_2 s_3 ^2 )/(c^2 c_2 c_3 ^2 ))=2(s_3 /(c^2 c_3 )) s_2 c^2 c_3 ^2 c_4 +s_4 c^2 c_2 c_3 ^2 −s^2 s_3 ^2 s_4 c_2 −s^2 s_2 s_3 ^2 c_4 =2s_3 c_2 c_3 c_4 (c^2 c_3 ^2 −s^2 s_3 ^2 )(s_2 c_4 +s_4 c_2 )=2s_3 c_2 c_3 c_4 (c^2 c_3 ^2 −(1−c^2 )(1−c_3 ^2 ))(s_2 c_4 +s_4 c_2 )=2s_3 c_2 c_3 c_4 (c^2 +c_3 ^2 −1)(s_2 c_4 +s_4 c_2 )=2s_3 c_2 c_3 c_4 now use trigonometric formulas until you reach this... ((1/2)(c_2 +c_6 ))(s_6 )=(1/4)(s_4 +s_8 +s_(12) ) ...and again: (1/4)(s_4 +s_8 +s_(12) )=(1/4)(s_4 +s_8 +s_(12) )
$$\mathrm{let}\:\mathrm{me}\:\mathrm{write}\:{c}_{{n}} /{s}_{{n}} /{t}_{{n}} \:\mathrm{for}\:\mathrm{cos}\:{nx}/\mathrm{sin}\:{nx}/\mathrm{tan}\:{nx} \\ $$$$\left({t}_{\mathrm{4}} +{t}_{\mathrm{2}} \right)\left(\mathrm{1}−{t}_{\mathrm{3}} ^{\mathrm{2}} {t}^{\mathrm{2}} \right)=\mathrm{2}\frac{{t}_{\mathrm{3}} }{{c}^{\mathrm{2}} } \\ $$$${t}_{\mathrm{2}} +{t}_{\mathrm{4}} −{t}^{\mathrm{2}} {t}_{\mathrm{3}} ^{\mathrm{2}} {t}_{\mathrm{4}} −{t}^{\mathrm{2}} {t}_{\mathrm{2}} {t}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{2}\frac{{t}_{\mathrm{3}} }{{c}^{\mathrm{2}} } \\ $$$$\frac{{s}_{\mathrm{2}} }{{c}_{\mathrm{2}} }+\frac{{s}_{\mathrm{4}} }{{c}_{\mathrm{4}} }−\frac{{s}^{\mathrm{2}} {s}_{\mathrm{3}} ^{\mathrm{2}} {s}_{\mathrm{4}} }{{c}^{\mathrm{2}} {c}_{\mathrm{3}} ^{\mathrm{2}} {c}_{\mathrm{4}} }−\frac{{s}^{\mathrm{2}} {s}_{\mathrm{2}} {s}_{\mathrm{3}} ^{\mathrm{2}} }{{c}^{\mathrm{2}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} ^{\mathrm{2}} }=\mathrm{2}\frac{{s}_{\mathrm{3}} }{{c}^{\mathrm{2}} {c}_{\mathrm{3}} } \\ $$$${s}_{\mathrm{2}} {c}^{\mathrm{2}} {c}_{\mathrm{3}} ^{\mathrm{2}} {c}_{\mathrm{4}} +{s}_{\mathrm{4}} {c}^{\mathrm{2}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} ^{\mathrm{2}} −{s}^{\mathrm{2}} {s}_{\mathrm{3}} ^{\mathrm{2}} {s}_{\mathrm{4}} {c}_{\mathrm{2}} −{s}^{\mathrm{2}} {s}_{\mathrm{2}} {s}_{\mathrm{3}} ^{\mathrm{2}} {c}_{\mathrm{4}} =\mathrm{2}{s}_{\mathrm{3}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} {c}_{\mathrm{4}} \\ $$$$\left({c}^{\mathrm{2}} {c}_{\mathrm{3}} ^{\mathrm{2}} −{s}^{\mathrm{2}} {s}_{\mathrm{3}} ^{\mathrm{2}} \right)\left({s}_{\mathrm{2}} {c}_{\mathrm{4}} +{s}_{\mathrm{4}} {c}_{\mathrm{2}} \right)=\mathrm{2}{s}_{\mathrm{3}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} {c}_{\mathrm{4}} \\ $$$$\left({c}^{\mathrm{2}} {c}_{\mathrm{3}} ^{\mathrm{2}} −\left(\mathrm{1}−{c}^{\mathrm{2}} \right)\left(\mathrm{1}−{c}_{\mathrm{3}} ^{\mathrm{2}} \right)\right)\left({s}_{\mathrm{2}} {c}_{\mathrm{4}} +{s}_{\mathrm{4}} {c}_{\mathrm{2}} \right)=\mathrm{2}{s}_{\mathrm{3}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} {c}_{\mathrm{4}} \\ $$$$\left({c}^{\mathrm{2}} +{c}_{\mathrm{3}} ^{\mathrm{2}} −\mathrm{1}\right)\left({s}_{\mathrm{2}} {c}_{\mathrm{4}} +{s}_{\mathrm{4}} {c}_{\mathrm{2}} \right)=\mathrm{2}{s}_{\mathrm{3}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} {c}_{\mathrm{4}} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\mathrm{until}\:\mathrm{you} \\ $$$$\mathrm{reach}\:\mathrm{this}… \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\left({c}_{\mathrm{2}} +{c}_{\mathrm{6}} \right)\right)\left({s}_{\mathrm{6}} \right)=\frac{\mathrm{1}}{\mathrm{4}}\left({s}_{\mathrm{4}} +{s}_{\mathrm{8}} +{s}_{\mathrm{12}} \right) \\ $$$$…\mathrm{and}\:\mathrm{again}: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({s}_{\mathrm{4}} +{s}_{\mathrm{8}} +{s}_{\mathrm{12}} \right)=\frac{\mathrm{1}}{\mathrm{4}}\left({s}_{\mathrm{4}} +{s}_{\mathrm{8}} +{s}_{\mathrm{12}} \right) \\ $$
Answered by MJS last updated on 05/Jul/18
t=arctan α tan α → t tan 2α → ((2t)/((1−t)(1+t))) tan 3α → ((t(3−t^2 ))/(1−3t^2 )) tan 4α → ((4t(1−t)(1+t))/((1+2t−t^2 )(1−2t−t^2 ))) sec^2 α → 1+t^2 left side (((2t)/((1−t)(1+t)))+((4t(1−t)(1+t))/((1+2t−t^2 )(1−2t−t^2 ))))= =((2t(1−3t^2 )(3−t^2 ))/((1+2t−t^2 )(1−2t−t^2 )(1−t)(1+t)))=A (1−(((t(3−t^2 ))/(1−3t^2 )))^2 t^2 )=(((1−6t^2 +t^4 )(1+t^2 )(1−t)(1+t))/((1−3t^2 )^2 ))=B A×B=((2t(3−t^2 )(1+t^2 ))/(1−3t^2 )) right side 2((t(3−t^2 ))/(1−3t^2 ))(1+t^2 )
$${t}=\mathrm{arctan}\:\alpha \\ $$$$\mathrm{tan}\:\alpha\:\rightarrow\:{t} \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha\:\rightarrow\:\frac{\mathrm{2}{t}}{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)} \\ $$$$\mathrm{tan}\:\mathrm{3}\alpha\:\rightarrow\:\frac{{t}\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\mathrm{4}\alpha\:\rightarrow\:\frac{\mathrm{4}{t}\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}{\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2}{t}−{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{sec}^{\mathrm{2}} \:\alpha\:\rightarrow\:\mathrm{1}+{t}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{left}\:\mathrm{side} \\ $$$$\left(\frac{\mathrm{2}{t}}{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}+\frac{\mathrm{4}{t}\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}{\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2}{t}−{t}^{\mathrm{2}} \right)}\right)= \\ $$$$=\frac{\mathrm{2}{t}\left(\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} \right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2}{t}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}={A} \\ $$$$\left(\mathrm{1}−\left(\frac{{t}\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} }\right)^{\mathrm{2}} {t}^{\mathrm{2}} \right)=\frac{\left(\mathrm{1}−\mathrm{6}{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}{\left(\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }={B} \\ $$$${A}×{B}=\frac{\mathrm{2}{t}\left(\mathrm{3}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{right}\:\mathrm{side} \\ $$$$\mathrm{2}\frac{{t}\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} }\left(\mathrm{1}+{t}^{\mathrm{2}} \right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jul/18
(((sin4α)/(cos4α))+((sin2α)/(cos2α)))(((cos^2 3αcos^2 α−sin^2 3αsin^2 α)/(cos^2 3αcos^2 α))) (((sin6α)/(cos4αcos2α)))(((((1+c6α)/2).((1+c2α)/2)−((1−c6α)/2).((1−c2α)/2))/(c^2 3α.c^2 α))) ((2s3αc3α)/(c4αc2α)).(1/4).((2c6α+2c2α)/(cos^2 3αcos^2 α)) ((sin3αcos3α)/(cos4αcos2α)).((cos6α+cos2α)/(cos^2 3αcos^2 α)) ((sin3αcos3α)/(cos4αcos2α)).2((cos4αcos2α)/(cos^2 3αcos^2 α)) 2tan3αsec^2 α
$$\left(\frac{{sin}\mathrm{4}\alpha}{{cos}\mathrm{4}\alpha}+\frac{{sin}\mathrm{2}\alpha}{{cos}\mathrm{2}\alpha}\right)\left(\frac{{cos}^{\mathrm{2}} \mathrm{3}\alpha{cos}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \mathrm{3}\alpha{sin}^{\mathrm{2}} \alpha}{{cos}^{\mathrm{2}} \mathrm{3}\alpha{cos}^{\mathrm{2}} \alpha}\right) \\ $$$$\left(\frac{{sin}\mathrm{6}\alpha}{{cos}\mathrm{4}\alpha{cos}\mathrm{2}\alpha}\right)\left(\frac{\frac{\mathrm{1}+{c}\mathrm{6}\alpha}{\mathrm{2}}.\frac{\mathrm{1}+{c}\mathrm{2}\alpha}{\mathrm{2}}−\frac{\mathrm{1}−{c}\mathrm{6}\alpha}{\mathrm{2}}.\frac{\mathrm{1}−{c}\mathrm{2}\alpha}{\mathrm{2}}}{{c}^{\mathrm{2}} \mathrm{3}\alpha.{c}^{\mathrm{2}} \alpha}\right) \\ $$$$ \\ $$$$\frac{\mathrm{2}{s}\mathrm{3}\alpha{c}\mathrm{3}\alpha}{{c}\mathrm{4}\alpha{c}\mathrm{2}\alpha}.\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{2}{c}\mathrm{6}\alpha+\mathrm{2}{c}\mathrm{2}\alpha}{{cos}^{\mathrm{2}} \mathrm{3}\alpha{cos}^{\mathrm{2}} \alpha} \\ $$$$\frac{{sin}\mathrm{3}\alpha{cos}\mathrm{3}\alpha}{{cos}\mathrm{4}\alpha{cos}\mathrm{2}\alpha}.\frac{{cos}\mathrm{6}\alpha+{cos}\mathrm{2}\alpha}{{cos}^{\mathrm{2}} \mathrm{3}\alpha{cos}^{\mathrm{2}} \alpha} \\ $$$$\frac{{sin}\mathrm{3}\alpha{cos}\mathrm{3}\alpha}{{cos}\mathrm{4}\alpha{cos}\mathrm{2}\alpha}.\mathrm{2}\frac{{cos}\mathrm{4}\alpha{cos}\mathrm{2}\alpha}{{cos}^{\mathrm{2}} \mathrm{3}\alpha{cos}^{\mathrm{2}} \alpha} \\ $$$$\mathrm{2}{tan}\mathrm{3}\alpha{sec}^{\mathrm{2}} \alpha \\ $$$$\:\:\:\:\: \\ $$
Answered by math1967 last updated on 05/Jul/18
(((tan3α+tanα)/(1−tan 3αtan α)) +((tan 3α−tan α)/(1+tan 3αtan α)))(1−tan^2 3αtan^2 α) ((2tan3α+2tan^2 αtan3α)/((1−tan^2 3αtan^2 α)))×(1−tan^2 3αtan^2 α) 2tan3α(1+tan^2 α)=2tan3αsec^2 α
$$\left(\frac{{tan}\mathrm{3}\alpha+{tan}\alpha}{\mathrm{1}−\mathrm{tan}\:\mathrm{3}\alpha\mathrm{tan}\:\alpha}\:+\frac{\mathrm{tan}\:\mathrm{3}\alpha−\mathrm{tan}\:\alpha}{\mathrm{1}+\mathrm{tan}\:\mathrm{3}\alpha\mathrm{tan}\:\alpha}\right)\left(\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}\alpha{tan}^{\mathrm{2}} \alpha\right) \\ $$$$\frac{\mathrm{2}{tan}\mathrm{3}\alpha+\mathrm{2}{tan}^{\mathrm{2}} \alpha{tan}\mathrm{3}\alpha}{\left(\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}\alpha{tan}^{\mathrm{2}} \alpha\right)}×\left(\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}\alpha{tan}^{\mathrm{2}} \alpha\right) \\ $$$$\mathrm{2}{tan}\mathrm{3}\alpha\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)=\mathrm{2}{tan}\mathrm{3}\alpha{sec}^{\mathrm{2}} \alpha \\ $$

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