Question Number 155466 by cortano last updated on 01/Oct/21
$$\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\ldots+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right)=? \\ $$
Commented by VIDDD last updated on 01/Oct/21
$${can}\:{u}\:{show}\:{u}\:{solution} \\ $$
Commented by VIDDD last updated on 01/Oct/21
$${I}\:\mathrm{wanna}\:\mathrm{see}\:\mathrm{ur}\:\mathrm{sulotion}\:\mathrm{plz}\: \\ $$
Answered by puissant last updated on 01/Oct/21
$${tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{12}}\right)+…+{tan}^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right) \\ $$$$=\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}−\mathrm{2}+\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}−\mathrm{2}+\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}−\mathrm{2}+{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}+\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}+\mathrm{3} \\ $$$$=\mathrm{4}\left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}+\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}\right)+\mathrm{3} \\ $$$$=\mathrm{4}\left(\frac{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right)+\mathrm{3} \\ $$$$=\frac{\mathrm{16}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}+\mathrm{3}\:=\:\mathrm{32}+\mathrm{3}\:=\:\mathrm{35}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\therefore\because\:\:{S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\sum}}{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{16}}\right)=\mathrm{35}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……………\mathscr{L}{e}\:{puissant}……………. \\ $$
Commented by VIDDD last updated on 01/Oct/21
$$\mathrm{thanks}\:\mathrm{sir}\heartsuit \\ $$
Commented by peter frank last updated on 01/Oct/21
$$\mathrm{great} \\ $$
Answered by john_santu last updated on 01/Oct/21
$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tan}^{\mathrm{2}} \:\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{2}}\right)=\:\frac{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\sum}}\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{16}}\right)=\frac{\mathrm{7}×\mathrm{15}}{\mathrm{3}}=\:\mathrm{35} \\ $$