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...](https://www.tinkutara.com/question/Q59841.png)
$${T}_{{r}} =\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{2}{r}−\mathrm{1}} }{\mathrm{2}{r}−\mathrm{1}} \\ $$$${l}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\Sigma\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{2}{r}−\mathrm{1}} }{\mathrm{2}{r}−\mathrm{1}} \\ $$$${l}=\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{1}} }{\mathrm{1}}+\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{5}} }{\mathrm{5}}+…\infty \\ $$$${now}\:{ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+…\infty \\ $$$${ln}\left(\mathrm{1}−{x}\right)=−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+…\infty \\ $$$${ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right) \\ $$$$=\mathrm{2}\left[{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+…\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$${so}\:{l}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{1}−{cos}\mathrm{2}\theta}\right) \\ $$$${l}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}{cos}^{\mathrm{2}} \theta}{\mathrm{2}{sin}^{\mathrm{2}} \theta}\right) \\ $$$${l}={lncot}\theta \\ $$$${when}\:{cot}\theta=\mathrm{1}\:\: \\ $$$${ln}\left({cot}\theta\right) \\ $$$${ln}\left(\mathrm{1}\right) \\ $$$$=\mathrm{0} \\ $$$$\left[{l}\right] \\ $$$$\left[\mathrm{0}\right] \\ $$$$=\mathrm{0} \\ $$$${when}\:{cot}\theta=\mathrm{2}\:\: \\ $$$${ln}\left({cot}\theta\right) \\ $$$${ln}\mathrm{2} \\ $$$$=\mathrm{0}.\mathrm{0}.\mathrm{6931} \\ $$$$\left[\mathrm{0}.\mathrm{6931}\right]=\mathrm{0} \\ $$$$ \\ $$$${i}\:{have}\:{just}\:{tried}…{pls}\:{check}… \\ $$
Commented by bhanukumarb2@gmail.com last updated on 15/May/19
$${correct}\:{sir}\:{nice}\:{approch}\:{igot}\:{another}\:{mthd}\:{also} \\ $$
Commented by tanmay last updated on 15/May/19
$${thank}\:{you}\:{sir}… \\ $$
Commented by bhanukumarb2@gmail.com last updated on 15/May/19
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Commented by tanmay last updated on 15/May/19
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Commented by bhanukumarb2@gmail.com last updated on 15/May/19
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Commented by tanmay last updated on 15/May/19
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Commented by bhanukumarb2@gmail.com last updated on 15/May/19
$${so}\:{nice}\:{of}\:{u}\:{thanks}\:{plz}\:{see}\:{my}/{previos} \\ $$$${and}\:{new}\:{doubt}\:{thanks} \\ $$