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Question Number 155466 by cortano last updated on 01/Oct/21

tan^2 ((π/(16)))+tan^2 (((2π)/(16)))+…+tan^2 (((7π)/(16)))=?

$$\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\ldots+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right)=? \\ $$

Commented by VIDDD last updated on 01/Oct/21

can u show u solution

$${can}\:{u}\:{show}\:{u}\:{solution} \\ $$

Commented by VIDDD last updated on 01/Oct/21

I wanna see ur sulotion plz

$${I}\:\mathrm{wanna}\:\mathrm{see}\:\mathrm{ur}\:\mathrm{sulotion}\:\mathrm{plz}\: \\ $$

Answered by puissant last updated on 01/Oct/21

tan^2 ((π/(16)))+tan^2 (((2π)/(12)))+...+tan^2 (((7π)/(16))) =(4/(sin^2 ((π/8))))−2+(4/(sin^2 ((π/4))))−2+(4/(sin^2 (((3π)/8))))−2+tan^2 ((π/4)) =(4/(sin^2 ((π/8))))+(4/(sin^2 (((3π)/8))))+3 =4((1/(sin^2 ((π/8))))+(1/(sin^2 (((3π)/8)))))+3 =4(((sin^2 ((π/8))+cos^2 ((π/8)))/(sin^2 ((π/8))cos^2 ((π/8)))))+3 =((16)/(sin^2 ((π/4))))+3 = 32+3 = 35.. ∴∵ S=Σ_(k=1) ^7 tan^2 (((kπ)/(16)))=35... ...............Le puissant................

$${tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{12}}\right)+...+{tan}^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right) \\ $$$$=\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}−\mathrm{2}+\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}−\mathrm{2}+\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}−\mathrm{2}+{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}+\frac{\mathrm{4}}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}+\mathrm{3} \\ $$$$=\mathrm{4}\left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}+\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}\right)+\mathrm{3} \\ $$$$=\mathrm{4}\left(\frac{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right){cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}\right)+\mathrm{3} \\ $$$$=\frac{\mathrm{16}}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}+\mathrm{3}\:=\:\mathrm{32}+\mathrm{3}\:=\:\mathrm{35}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\therefore\because\:\:{S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\sum}}{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{16}}\right)=\mathrm{35}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...............\mathscr{L}{e}\:{puissant}................ \\ $$

Commented by VIDDD last updated on 01/Oct/21

thanks sir♥

$$\mathrm{thanks}\:\mathrm{sir}\heartsuit \\ $$

Commented by peter frank last updated on 01/Oct/21

great

$$\mathrm{great} \\ $$

Answered by john_santu last updated on 01/Oct/21

Σ_(k=1) ^n tan^2 (((kπ)/(2n+2)))= ((n(2n+1))/3) Σ_(k=1) ^7 tan^2 (((kπ)/(16)))=((7×15)/3)= 35

$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tan}^{\mathrm{2}} \:\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{2}}\right)=\:\frac{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\sum}}\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{16}}\right)=\frac{\mathrm{7}×\mathrm{15}}{\mathrm{3}}=\:\mathrm{35} \\ $$

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