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Question-59835




Question Number 59835 by bhanukumarb2@gmail.com last updated on 15/May/19
Answered by tanmay last updated on 15/May/19
T_r =(((cos2θ)^(2r−1) )/(2r−1))  l=lim_(n→∞)  Σ(((cos2θ)^(2r−1) )/(2r−1))  l=(((cos2θ)^1 )/1)+(((cos2θ)^3 )/3)+(((cos2θ)^5 )/5)+...∞  now ln(1+x)=x−(x^2 /2)+(x^3 /3)−(x^4 /4)+...∞  ln(1−x)=−x−(x^2 /2)−(x^3 /3)−(x^4 /4)+...∞  ln(1+x)−ln(1−x)  =2[x+(x^3 /3)+(x^5 /5)+...]  (1/2)ln(((1+x)/(1−x)))  so l=(1/2)ln(((1+cos2θ)/(1−cos2θ)))  l=(1/2)ln(((2cos^2 θ)/(2sin^2 θ)))  l=lncotθ  when cotθ=1    ln(cotθ)  ln(1)  =0  [l]  [0]  =0  when cotθ=2    ln(cotθ)  ln2  =0.0.6931  [0.6931]=0    i have just tried...pls check...
$${T}_{{r}} =\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{2}{r}−\mathrm{1}} }{\mathrm{2}{r}−\mathrm{1}} \\ $$$${l}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\Sigma\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{2}{r}−\mathrm{1}} }{\mathrm{2}{r}−\mathrm{1}} \\ $$$${l}=\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{1}} }{\mathrm{1}}+\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{5}} }{\mathrm{5}}+…\infty \\ $$$${now}\:{ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+…\infty \\ $$$${ln}\left(\mathrm{1}−{x}\right)=−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+…\infty \\ $$$${ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right) \\ $$$$=\mathrm{2}\left[{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+…\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$${so}\:{l}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{1}−{cos}\mathrm{2}\theta}\right) \\ $$$${l}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}{cos}^{\mathrm{2}} \theta}{\mathrm{2}{sin}^{\mathrm{2}} \theta}\right) \\ $$$${l}={lncot}\theta \\ $$$${when}\:{cot}\theta=\mathrm{1}\:\: \\ $$$${ln}\left({cot}\theta\right) \\ $$$${ln}\left(\mathrm{1}\right) \\ $$$$=\mathrm{0} \\ $$$$\left[{l}\right] \\ $$$$\left[\mathrm{0}\right] \\ $$$$=\mathrm{0} \\ $$$${when}\:{cot}\theta=\mathrm{2}\:\: \\ $$$${ln}\left({cot}\theta\right) \\ $$$${ln}\mathrm{2} \\ $$$$=\mathrm{0}.\mathrm{0}.\mathrm{6931} \\ $$$$\left[\mathrm{0}.\mathrm{6931}\right]=\mathrm{0} \\ $$$$ \\ $$$${i}\:{have}\:{just}\:{tried}…{pls}\:{check}… \\ $$
Commented by bhanukumarb2@gmail.com last updated on 15/May/19
correct sir nice approch igot another mthd also
$${correct}\:{sir}\:{nice}\:{approch}\:{igot}\:{another}\:{mthd}\:{also} \\ $$
Commented by tanmay last updated on 15/May/19
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$
Commented by bhanukumarb2@gmail.com last updated on 15/May/19
sir are you jee aspirant
$${sir}\:{are}\:{you}\:{jee}\:{aspirant}\:\: \\ $$
Commented by tanmay last updated on 15/May/19
no i am 49 year old indian stay at nagpur  interest in physics and math
$${no}\:{i}\:{am}\:\mathrm{49}\:{year}\:{old}\:{indian}\:{stay}\:{at}\:{nagpur} \\ $$$${interest}\:{in}\:{physics}\:{and}\:{math} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 15/May/19
thats really nice u have really very  indepth knowledge i m actully jee aspirant  sir please guide me and my doubts   thanks
$${thats}\:{really}\:{nice}\:{u}\:{have}\:{really}\:{very} \\ $$$${indepth}\:{knowledge}\:{i}\:{m}\:{actully}\:{jee}\:{aspirant} \\ $$$${sir}\:{please}\:{guide}\:{me}\:{and}\:{my}\:{doubts}\: \\ $$$${thanks} \\ $$
Commented by tanmay last updated on 15/May/19
i this platform several think tank are present  they are superb i math and physis  example..MJS sir  MRW sir AJFOUR sir  Mr.Mathmaxup and Mr. Behi...  all are excellent...we are here to give food to  our brain...
$${i}\:{this}\:{platform}\:{several}\:{think}\:{tank}\:{are}\:{present} \\ $$$${they}\:{are}\:{superb}\:{i}\:{math}\:{and}\:{physis} \\ $$$${example}..{MJS}\:{sir}\:\:{MRW}\:{sir}\:{AJFOUR}\:{sir} \\ $$$${Mr}.{Mathmaxup}\:{and}\:{Mr}.\:{Behi}… \\ $$$${all}\:{are}\:{excellent}…{we}\:{are}\:{here}\:{to}\:{give}\:{food}\:{to} \\ $$$${our}\:{brain}… \\ $$
Commented by bhanukumarb2@gmail.com last updated on 15/May/19
so nice of u thanks plz see my/previos  and new doubt thanks
$${so}\:{nice}\:{of}\:{u}\:{thanks}\:{plz}\:{see}\:{my}/{previos} \\ $$$${and}\:{new}\:{doubt}\:{thanks} \\ $$

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