Question Number 57373 by gunawan last updated on 03/Apr/19
$$\mathrm{tan}\:\mathrm{1}°+\mathrm{tan}\:\mathrm{5}°+\mathrm{tan}\:\mathrm{9}°+\ldots+\mathrm{tan}\:\mathrm{177}°=... \\ $$
Commented by mr W last updated on 03/Apr/19
$$=\mathrm{45} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19
![let y=cosxcos5xcos9x...cos(x+(n−1)4x) y=cosxcos5xcos9x...cos(4nx−3x) lny=lncosx+lncos5x+lncos9x+..+lncos(4nx−3x3 (1/y)×(dy/dx)=−[(tanx)+tsn5x+tan9x+..tan(4nx−3x)] this method applicable if we know value of y](Q57402.png)
$${let} \\ $$$${y}={cosxcos}\mathrm{5}{xcos}\mathrm{9}{x}...{cos}\left({x}+\left({n}−\mathrm{1}\right)\mathrm{4}{x}\right) \\ $$$${y}={cosxcos}\mathrm{5}{xcos}\mathrm{9}{x}...{cos}\left(\mathrm{4}{nx}−\mathrm{3}{x}\right) \\ $$$${lny}={lncosx}+{lncos}\mathrm{5}{x}+{lncos}\mathrm{9}{x}+..+{lncos}\left(\mathrm{4}{nx}−\mathrm{3}{x}\mathrm{3}\right. \\ $$$$\frac{\mathrm{1}}{{y}}×\frac{{dy}}{{dx}}=−\left[\left({tanx}\right)+{tsn}\mathrm{5}{x}+{tan}\mathrm{9}{x}+..{tan}\left(\mathrm{4}{nx}−\mathrm{3}{x}\right)\right] \\ $$$${this}\:{method}\:{applicable}\:{if}\:{we}\:{know}\:{value}\:{of}\:{y} \\ $$
Commented by mr W last updated on 03/Apr/19
$${it}'{s}\:{wrong}\:{sir}.\:{please}\:{check}: \\ $$$$\frac{{d}\left(\mathrm{ln}\:\mathrm{cos}\:{nx}\right)}{{dx}}=−{n}\:\mathrm{tan}\:{nx}\:\neq−\mathrm{tan}\:{nx} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19
$${yes}\:{yes}\:...{you}\:{are}\:{correct}..{i}\:{am}\:{just}\:{giving}\:{example}.. \\ $$$${let}\:{me}\:{give}\:{real}\:{problem}\:{example}.. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19
