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If $z=sqrt3+i=(a+ib)(c+id)$, then find $tan^-1dfracba+tan^-1dfracdc$
Find area bounded by two unequal chords and an arc in a discProve $left|fracz_1z_2right|=fracz_1$ for two complex numbersShow that $y=frac4sintheta2+costheta-theta$ is increasing function when $theta in [0,fracpi2]$Proving a proposition involving the argument of a complex numberFind the value of $4tan(theta _1)tan(theta _2)$ if $theta _1$ and $theta _2$ are the solutions of the given equationVerify $arccosx+arcsinx=fracpi2$Trouble with Trig Substitution IntegralEvaluation in a complex functionIs there any formula to find $sin^-1a +sin^-1b$If $z=dfrac(z_1+barz_2)z_1z_2barz_1$ where $z_1=1+2i$ and $z_2=1-i$, then find $arg(z)$
$begingroup$
If $z=sqrt3+i=(a+ib)(c+id)$, then $tan^-1dfracba+tan^-1dfracdc=$ _______________
$$
arg z=theta,quadarg(a+ib)=theta_1,quadarg(c+id)=theta_2impliesarg(z)=theta=theta_1+theta_2=fracpi6\
tantheta=tanfracpi6=frac1sqrt3,quadtantheta_1=fracba,quadtantheta_2=fracdc\
impliestheta_1=npi+tan^-1fracba,quadtheta_2=mpi+tan^-1fracdc\
theta=fracpi6=(n+m)pi+tan^-1fracba+tan^-1fracdc\
tan^-1fracba+tan^-1fracdc=fracpi6-(n+m)pi=fracpi6+kpi
$$
Is it the right way to prove that $tan^-1dfracba+tan^-1dfracdc=npi+dfracpi6$ ?
trigonometry complex-numbers
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
$endgroup$
add a comment |
$begingroup$
If $z=sqrt3+i=(a+ib)(c+id)$, then $tan^-1dfracba+tan^-1dfracdc=$ _______________
$$
arg z=theta,quadarg(a+ib)=theta_1,quadarg(c+id)=theta_2impliesarg(z)=theta=theta_1+theta_2=fracpi6\
tantheta=tanfracpi6=frac1sqrt3,quadtantheta_1=fracba,quadtantheta_2=fracdc\
impliestheta_1=npi+tan^-1fracba,quadtheta_2=mpi+tan^-1fracdc\
theta=fracpi6=(n+m)pi+tan^-1fracba+tan^-1fracdc\
tan^-1fracba+tan^-1fracdc=fracpi6-(n+m)pi=fracpi6+kpi
$$
Is it the right way to prove that $tan^-1dfracba+tan^-1dfracdc=npi+dfracpi6$ ?
trigonometry complex-numbers
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
$endgroup$
add a comment |
$begingroup$
If $z=sqrt3+i=(a+ib)(c+id)$, then $tan^-1dfracba+tan^-1dfracdc=$ _______________
$$
arg z=theta,quadarg(a+ib)=theta_1,quadarg(c+id)=theta_2impliesarg(z)=theta=theta_1+theta_2=fracpi6\
tantheta=tanfracpi6=frac1sqrt3,quadtantheta_1=fracba,quadtantheta_2=fracdc\
impliestheta_1=npi+tan^-1fracba,quadtheta_2=mpi+tan^-1fracdc\
theta=fracpi6=(n+m)pi+tan^-1fracba+tan^-1fracdc\
tan^-1fracba+tan^-1fracdc=fracpi6-(n+m)pi=fracpi6+kpi
$$
Is it the right way to prove that $tan^-1dfracba+tan^-1dfracdc=npi+dfracpi6$ ?
trigonometry complex-numbers
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
$endgroup$
If $z=sqrt3+i=(a+ib)(c+id)$, then $tan^-1dfracba+tan^-1dfracdc=$ _______________
$$
arg z=theta,quadarg(a+ib)=theta_1,quadarg(c+id)=theta_2impliesarg(z)=theta=theta_1+theta_2=fracpi6\
tantheta=tanfracpi6=frac1sqrt3,quadtantheta_1=fracba,quadtantheta_2=fracdc\
impliestheta_1=npi+tan^-1fracba,quadtheta_2=mpi+tan^-1fracdc\
theta=fracpi6=(n+m)pi+tan^-1fracba+tan^-1fracdc\
tan^-1fracba+tan^-1fracdc=fracpi6-(n+m)pi=fracpi6+kpi
$$
Is it the right way to prove that $tan^-1dfracba+tan^-1dfracdc=npi+dfracpi6$ ?
trigonometry complex-numbers
trigonometry complex-numbers
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
edited Mar 14 at 12:56
ss1729
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
asked Mar 14 at 7:49
ss1729ss1729
2,03711124
2,03711124
add a comment |
add a comment |
1 Answer
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$begingroup$
I think that a simpler way would bebeginaligntanleft(arctanleft(frac baright)+arctanleft(frac dcright)right)&=fracfrac ba+frac dc1-frac batimesfrac dc\&=fracbc+adac-bd\&=frac1sqrt3.endalign
edited Mar 14 at 8:00
ss1729
2,03711124
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
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$begingroup$
I think that a simpler way would bebeginaligntanleft(arctanleft(frac baright)+arctanleft(frac dcright)right)&=fracfrac ba+frac dc1-frac batimesfrac dc\&=fracbc+adac-bd\&=frac1sqrt3.endalign
edited Mar 14 at 8:00
ss1729
2,03711124
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
I think that a simpler way would bebeginaligntanleft(arctanleft(frac baright)+arctanleft(frac dcright)right)&=fracfrac ba+frac dc1-frac batimesfrac dc\&=fracbc+adac-bd\&=frac1sqrt3.endalign
edited Mar 14 at 8:00
ss1729
2,03711124
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
I think that a simpler way would bebeginaligntanleft(arctanleft(frac baright)+arctanleft(frac dcright)right)&=fracfrac ba+frac dc1-frac batimesfrac dc\&=fracbc+adac-bd\&=frac1sqrt3.endalign
edited Mar 14 at 8:00
ss1729
2,03711124
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
I think that a simpler way would bebeginaligntanleft(arctanleft(frac baright)+arctanleft(frac dcright)right)&=fracfrac ba+frac dc1-frac batimesfrac dc\&=fracbc+adac-bd\&=frac1sqrt3.endalign
edited Mar 14 at 8:00
ss1729
2,03711124
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
edited Mar 14 at 8:00
ss1729
2,03711124
edited Mar 14 at 8:00
ss1729
2,03711124
edited Mar 14 at 8:00
ss1729
2,03711124
2,03711124
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
answered Mar 14 at 7:55
José Carlos SantosJosé Carlos Santos
169k23132237
169k23132237
add a comment |
add a comment |
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