Question Number 14152 by tawa tawa last updated on 28/May/17
Answered by ajfour last updated on 29/May/17
![L=lim_(θ→π/6) [((tan^2 θ+tan^4 θ+tan^6 θ+...)/((1+tan^2 θ−(5/6))+(1+tan^2 θ−(5/6))^2 +...))] L=(((1/3)+((1/3))^2 +((1/3))^3 +...)/((1/2)+((1/2))^2 +((1/2))^3 +...)) L=(((1/3)((1/(1−1/3))))/((1/2)((1/(1−1/2)))))=(((1/3)×(3/2))/((1/2)×(2/1))) = (1/2) .](Q14178.png)
$${L}=\underset{\theta\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\:\left[\frac{\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{tan}\:^{\mathrm{4}} \theta+\mathrm{tan}\:^{\mathrm{6}} \theta+...}{\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta−\frac{\mathrm{5}}{\mathrm{6}}\right)+\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta−\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}} +...}\right] \\ $$$$\:\:\:\:\:{L}=\frac{\frac{\mathrm{1}}{\mathrm{3}}+\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} +...}{\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} +...} \\ $$$$\:\:\:\:\:\:{L}=\frac{\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{1}/\mathrm{3}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{1}/\mathrm{2}}\right)}=\frac{\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{1}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$$ \\ $$
Commented by tawa tawa last updated on 29/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$