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Question Number 38495 by kunal1234523 last updated on 26/Jun/18

prove that tan 3a tan 2a tan a = tan 3a − tan 2a − tan a

$${prove}\:{that} \\ $$$$\boldsymbol{\mathrm{tan}}\:\mathrm{3}\boldsymbol{{a}}\:\boldsymbol{\mathrm{tan}}\:\mathrm{2}\boldsymbol{{a}}\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{{a}}\:=\:\:\boldsymbol{\mathrm{tan}}\:\mathrm{3}\boldsymbol{{a}}\:−\:\boldsymbol{\mathrm{tan}}\:\mathrm{2}\boldsymbol{{a}}\:−\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{{a}} \\ $$

Answered by kunal1234523 last updated on 26/Jun/18

tan 3a = tan (a+2a) ⇒tan 3a = ((tan a + tan 2a)/(1 − tan a tan 2a)) ⇒tan 3a − tan a tan 2a tan 3a = tan a + tan 2a ⇒tan 3a tan 2a tan a = tan 3a − tan 2a −tan a

$${tan}\:\mathrm{3}{a}\:=\:{tan}\:\left({a}+\mathrm{2}{a}\right) \\ $$$$\Rightarrow{tan}\:\mathrm{3}{a}\:=\:\frac{{tan}\:{a}\:+\:{tan}\:\mathrm{2}{a}}{\mathrm{1}\:−\:{tan}\:{a}\:{tan}\:\mathrm{2}{a}} \\ $$$$\Rightarrow{tan}\:\mathrm{3}{a}\:−\:{tan}\:{a}\:{tan}\:\mathrm{2}{a}\:{tan}\:\mathrm{3}{a}\:=\:{tan}\:{a}\:+\:{tan}\:\mathrm{2}{a} \\ $$$$\Rightarrow{tan}\:\mathrm{3}{a}\:{tan}\:\mathrm{2}{a}\:{tan}\:{a}\:=\:{tan}\:\mathrm{3}{a}\:−\:{tan}\:\mathrm{2}{a}\:−{tan}\:{a} \\ $$

Answered by kunal1234523 last updated on 26/Jun/18

tan 2a = tan (3a − a) ⇒tan 2a = ((tan 3a − tan a)/(1 + tan 3a tan a)) ⇒tan 2a + tan 3a tan 2a tan a = tan 3a − tan a ⇒tan 3a tan 2a tan a = tan 3a − tan2a − tan a also tan a = tan (3a − 2a) ⇒tan a = ((tan 3a − tan 2a)/(1 + tan 3a tan 2a)) ⇒tan a + tan 3a tan 2a tan a = tan 3a − tan 2a ⇒tan 3a tan 2a tan a = tan 3a − tan2a − tan a

$${tan}\:\mathrm{2}{a}\:=\:{tan}\:\left(\mathrm{3}{a}\:−\:{a}\right) \\ $$$$\Rightarrow{tan}\:\mathrm{2}{a}\:=\:\frac{{tan}\:\mathrm{3}{a}\:−\:{tan}\:{a}}{\mathrm{1}\:+\:{tan}\:\mathrm{3}{a}\:{tan}\:{a}} \\ $$$$\Rightarrow{tan}\:\mathrm{2}{a}\:+\:{tan}\:\mathrm{3}{a}\:{tan}\:\mathrm{2}{a}\:{tan}\:{a}\:=\:{tan}\:\mathrm{3}{a}\:−\:{tan}\:{a} \\ $$$$\Rightarrow{tan}\:\mathrm{3}{a}\:{tan}\:\mathrm{2}{a}\:{tan}\:{a}\:=\:{tan}\:\mathrm{3}{a}\:−\:{tan}\mathrm{2}{a}\:−\:{tan}\:{a} \\ $$$$\mathrm{also} \\ $$$${tan}\:{a}\:=\:{tan}\:\left(\mathrm{3}{a}\:−\:\mathrm{2}{a}\right) \\ $$$$\Rightarrow{tan}\:{a}\:=\:\frac{{tan}\:\mathrm{3}{a}\:−\:{tan}\:\mathrm{2}{a}}{\mathrm{1}\:+\:{tan}\:\mathrm{3}{a}\:{tan}\:\mathrm{2}{a}} \\ $$$$\Rightarrow{tan}\:{a}\:+\:{tan}\:\mathrm{3}{a}\:{tan}\:\mathrm{2}{a}\:{tan}\:{a}\:=\:{tan}\:\mathrm{3}{a}\:−\:{tan}\:\mathrm{2}{a} \\ $$$$\Rightarrow{tan}\:\mathrm{3}{a}\:{tan}\:\mathrm{2}{a}\:{tan}\:{a}\:=\:{tan}\:\mathrm{3}{a}\:−\:{tan}\mathrm{2}{a}\:−\:{tan}\:{a} \\ $$

Answered by kunal1234523 last updated on 26/Jun/18

but there should be one more apporach and I am stucked there LHS ((sin 3a sin 2a sin a)/(cos 3a cos 2a cos a)) =((sin 3a (cos a − cos 3a))/(2 cos 3a cos 2a cos a)) =((sin 3a cos a)/(2cos 3a cos 2a cos a)) − ((sin 3a cos3a)/(2 cos 3a cos 2a cos a)) =((tan 3a)/(2 cos 2a)) − ((sin(a + 2a))/(2 cos a cos 2a)) =((tan 3a)/(2 cos 2a)) − (((sin a cos 2a)/(2 cos a cos 2a)) + ((cos a sin 2a)/(2 cos a cos 2a))) =((tan 3a)/(2 cos 2a))− ((tan 2a)/2) − ((tan a)/2) =(1/2)[((tan 3a)/(cos 2a)) − tan 2a − tan a] now what should I do.....

$${but}\:{there}\:{should}\:{be}\:{one}\:{more}\:{apporach}\:{and}\: \\ $$$${I}\:{am}\:{stucked}\:{there} \\ $$$${LHS} \\ $$$$\frac{{sin}\:\mathrm{3}{a}\:{sin}\:\mathrm{2}{a}\:{sin}\:{a}}{{cos}\:\mathrm{3}{a}\:{cos}\:\mathrm{2}{a}\:{cos}\:{a}}\: \\ $$$$=\frac{{sin}\:\mathrm{3}{a}\:\left({cos}\:{a}\:−\:{cos}\:\mathrm{3}{a}\right)}{\mathrm{2}\:{cos}\:\mathrm{3}{a}\:{cos}\:\mathrm{2}{a}\:{cos}\:{a}} \\ $$$$=\frac{{sin}\:\mathrm{3}{a}\:{cos}\:{a}}{\mathrm{2}{cos}\:\mathrm{3}{a}\:{cos}\:\mathrm{2}{a}\:{cos}\:{a}}\:−\:\frac{{sin}\:\mathrm{3}{a}\:{cos}\mathrm{3}{a}}{\mathrm{2}\:{cos}\:\mathrm{3}{a}\:{cos}\:\mathrm{2}{a}\:{cos}\:{a}} \\ $$$$=\frac{{tan}\:\mathrm{3}{a}}{\mathrm{2}\:{cos}\:\mathrm{2}{a}}\:−\:\frac{{sin}\left({a}\:+\:\mathrm{2}{a}\right)}{\mathrm{2}\:{cos}\:{a}\:{cos}\:\mathrm{2}{a}} \\ $$$$=\frac{{tan}\:\mathrm{3}{a}}{\mathrm{2}\:{cos}\:\mathrm{2}{a}}\:−\:\left(\frac{{sin}\:{a}\:{cos}\:\mathrm{2}{a}}{\mathrm{2}\:{cos}\:{a}\:{cos}\:\mathrm{2}{a}}\:+\:\frac{{cos}\:{a}\:{sin}\:\mathrm{2}{a}}{\mathrm{2}\:{cos}\:{a}\:{cos}\:\mathrm{2}{a}}\right) \\ $$$$=\frac{{tan}\:\mathrm{3}{a}}{\mathrm{2}\:{cos}\:\mathrm{2}{a}}−\:\frac{{tan}\:\mathrm{2}{a}}{\mathrm{2}}\:−\:\frac{{tan}\:{a}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{tan}\:\mathrm{3}{a}}{{cos}\:\mathrm{2}{a}}\:−\:{tan}\:\mathrm{2}{a}\:−\:{tan}\:{a}\right] \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{what}}\:\boldsymbol{{should}}\:\boldsymbol{{I}}\:\boldsymbol{{do}}..... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jun/18

scanning it meticulously...give time ...

$${scanning}\:{it}\:{meticulously}...{give}\:{time}\:... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jun/18

((tan3a)/(2cos2a))−((tan2a)/2)−((tana)/2) tan3a−tan2a−tana+((tan3a)/(2cos2a))−tan3a+tan2a −((tan2a)/2)+tana−((tana)/2) tan3a−tan2a−tana+tan3a((1/(2cos2a))−1)+(1/2) (tan2a+tana) do+tan3a((1/(2cos2a))−1)+(1/2)(((sin3a)/(cos2acosa))) =do+((sin3a)/(2cos3acos2a))−((sin3a)/(cos3a))+((sin3a)/(2cos2acosa)) do+sin3a((1/(2cos3acos2a))−(1/(cos3a))+(1/(2cos2acosa))) do+sin3a(((cosa+cos3a)/(2cos3acos2acosa))−(1/(cos3a))) do+sin3a(((2cos2a.cosa)/(2cos3acos2acosa))−(1/(cos3a))) do+sin3a((1/(cos3a))−(1/(cos3a))) =do +0 =tan3a−tan2a−tana proved

$$\frac{{tan}\mathrm{3}{a}}{\mathrm{2}{cos}\mathrm{2}{a}}−\frac{{tan}\mathrm{2}{a}}{\mathrm{2}}−\frac{{tana}}{\mathrm{2}} \\ $$$${tan}\mathrm{3}{a}−{tan}\mathrm{2}{a}−{tana}+\frac{{tan}\mathrm{3}{a}}{\mathrm{2}{cos}\mathrm{2}{a}}−{tan}\mathrm{3}{a}+{tan}\mathrm{2}{a} \\ $$$$−\frac{{tan}\mathrm{2}{a}}{\mathrm{2}}+{tana}−\frac{{tana}}{\mathrm{2}} \\ $$$${tan}\mathrm{3}{a}−{tan}\mathrm{2}{a}−{tana}+{tan}\mathrm{3}{a}\left(\frac{\mathrm{1}}{\mathrm{2}{cos}\mathrm{2}{a}}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({tan}\mathrm{2}{a}+{tana}\right) \\ $$$${do}+{tan}\mathrm{3}{a}\left(\frac{\mathrm{1}}{\mathrm{2}{cos}\mathrm{2}{a}}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{sin}\mathrm{3}{a}}{{cos}\mathrm{2}{acosa}}\right) \\ $$$$={do}+\frac{{sin}\mathrm{3}{a}}{\mathrm{2}{cos}\mathrm{3}{acos}\mathrm{2}{a}}−\frac{{sin}\mathrm{3}{a}}{{cos}\mathrm{3}{a}}+\frac{{sin}\mathrm{3}{a}}{\mathrm{2}{cos}\mathrm{2}{acosa}} \\ $$$${do}+{sin}\mathrm{3}{a}\left(\frac{\mathrm{1}}{\mathrm{2}{cos}\mathrm{3}{acos}\mathrm{2}{a}}−\frac{\mathrm{1}}{{cos}\mathrm{3}{a}}+\frac{\mathrm{1}}{\mathrm{2}{cos}\mathrm{2}{acosa}}\right) \\ $$$${do}+{sin}\mathrm{3}{a}\left(\frac{{cosa}+{cos}\mathrm{3}{a}}{\mathrm{2}{cos}\mathrm{3}{acos}\mathrm{2}{acosa}}−\frac{\mathrm{1}}{{cos}\mathrm{3}{a}}\right) \\ $$$${do}+{sin}\mathrm{3}{a}\left(\frac{\mathrm{2}{cos}\mathrm{2}{a}.{cosa}}{\mathrm{2}{cos}\mathrm{3}{acos}\mathrm{2}{acosa}}−\frac{\mathrm{1}}{{cos}\mathrm{3}{a}}\right) \\ $$$${do}+{sin}\mathrm{3}{a}\left(\frac{\mathrm{1}}{{cos}\mathrm{3}{a}}−\frac{\mathrm{1}}{{cos}\mathrm{3}{a}}\right) \\ $$$$={do}\:\:+\mathrm{0} \\ $$$$={tan}\mathrm{3}{a}−{tan}\mathrm{2}{a}−{tana}\:{proved} \\ $$$$ \\ $$$$ \\ $$

Commented by kunal1234523 last updated on 27/Jun/18

wow you take a “do” then proved everything else is zero smart.

$${wow}\:{you}\:{take}\:{a}\:``{do}''\:{then}\:{proved}\:{everything} \\ $$$${else}\:{is}\:{zero}\:{smart}. \\ $$

Answered by MJS last updated on 26/Jun/18

1=((tan 3α −tan 2α −tan α)/(tan 3α tan 2α tan α)) 1=(1/(tan 2α tan α))−(1/(tan 3α tan α))−(1/(tan 3α tan 2α)) α=arctan t 1=(1/(((2t)/(1−t^2 ))×t))−(1/(((3t−t^3 )/(3t^2 −1))×t))−(1/(((2t)/(1−t^2 ))×((3t−t^3 )/(3t^2 −1)))) 1=((1−t^2 )/(2t^2 ))−((3t^2 −1)/(t^2 (t^2 −3)))−(((1−t^2 )(3t^2 −1))/(2t^2 (t^2 −3))) 1=(((1−t^2 )(t^2 −3)−2(3t^2 −1)−(1−t^2 )(3t^2 −1))/(2t^2 (t^2 −3))) 1=(((1−t^2 )(t^2 −3)−(3t^2 −1)(2+1−t^2 ))/(2t^2 (t^2 −3))) 1=(((1−t^2 )(t^2 −3)−(3t^2 −1)(3−t^2 ))/(2t^2 (t^2 −3))) 1=(((t^2 −3)((1−t^2 )+(3t^2 −1)))/(2t^2 (t^2 −3))) 1=(((t^2 −3)2t^2 )/(2t^2 (t^2 −3))) 1=1 proved

$$ \\ $$$$\mathrm{1}=\frac{\mathrm{tan}\:\mathrm{3}\alpha\:−\mathrm{tan}\:\mathrm{2}\alpha\:−\mathrm{tan}\:\alpha}{\mathrm{tan}\:\mathrm{3}\alpha\:\mathrm{tan}\:\mathrm{2}\alpha\:\mathrm{tan}\:\alpha} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\alpha\:\mathrm{tan}\:\alpha}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{3}\alpha\:\mathrm{tan}\:\alpha}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{3}\alpha\:\mathrm{tan}\:\mathrm{2}\alpha} \\ $$$$\alpha=\mathrm{arctan}\:{t} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }×{t}}−\frac{\mathrm{1}}{\frac{\mathrm{3}{t}−{t}^{\mathrm{3}} }{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}×{t}}−\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }×\frac{\mathrm{3}{t}−{t}^{\mathrm{3}} }{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\mathrm{1}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}^{\mathrm{2}} }−\frac{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)}−\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)} \\ $$$$\mathrm{1}=\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} −\mathrm{3}\right)−\mathrm{2}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)} \\ $$$$\mathrm{1}=\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} −\mathrm{3}\right)−\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}+\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)} \\ $$$$\mathrm{1}=\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} −\mathrm{3}\right)−\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)} \\ $$$$\mathrm{1}=\frac{\left({t}^{\mathrm{2}} −\mathrm{3}\right)\left(\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)\right)}{\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)} \\ $$$$\mathrm{1}=\frac{\left({t}^{\mathrm{2}} −\mathrm{3}\right)\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)} \\ $$$$\mathrm{1}=\mathrm{1}\:\mathrm{proved} \\ $$

Commented by kunal1234523 last updated on 27/Jun/18

really simple and cool. there you had been taken t = tan α , i think

$${really}\:{simple}\:{and}\:{cool}.\:{there}\:{you}\:{had}\:{been}\: \\ $$$${taken}\:{t}\:=\:{tan}\:\alpha\:,\:{i}\:{think} \\ $$

Commented by MJS last updated on 27/Jun/18

yes.

$$\mathrm{yes}. \\ $$

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