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Question Number 132205 by Arijit last updated on 12/Feb/21

Commented by Arijit last updated on 12/Feb/21

$$\mathbf{Please}\mathbf{Help}\mathbf{me}\mathbf{to}\mathbf{solve}\mathbf{this}.....$$

Answered by mathmax by abdo last updated on 13/Feb/21

$$\mathrm{A}={\tan }^2\left(\frac{\pi }{16}\right)+{\tan }^2\left(\frac{7\pi }{16}\right)+{\tan }^2\left(\frac{2\pi }{16}\right)+{\tan }^2\left(\frac{6\pi }{16}\right)+{\tan }^2\left(\frac{3\pi }{16}\right)+{\tan }^2\left(\frac{5\pi }{16}\right)$$$$+{\tan }^2\left(\frac{4\pi }{16}\right)={\tan }^2\left(\frac{\pi }{16}\right)+{\tan }^2(\frac{\pi }{2}-\frac{\pi }{16})+{\tan }^2\left(\frac{2\pi }{16}\right)+{\tan }^2(\frac{\pi }{2}-\frac{2\pi }{16})$$$$+{\tan }^2\left(\frac{3\pi }{16}\right)+{\tan }^2(\frac{\pi }{2}-\frac{3\pi }{16})+{\tan }^2\left(\frac{\pi }{4}\right)$$$$={\tan }^2\left(\frac{\pi }{16}\right)+\frac{1}{{\tan }^2\left(\frac{\pi }{16}\right)}+{\tan }^2\left(\frac{2\pi }{16}\right)+\frac{1}{{\tan }^2\left(\frac{2\pi }{16}\right)}+{\tan }^2\left(\frac{3\pi }{16}\right)+\frac{1}{{\tan }^2\left(\frac{3\pi }{16}\right)}$$$$={\tan }^2\left(\frac{\pi }{16}\right)+\frac{1}{{\tan }^2\left(\frac{\pi }{16}\right)}+{(\sqrt{2}-1)}^2+\frac{1}{{(\sqrt{2}-1)}^2}+{\tan }^2\left(\frac{3\pi }{16}\right)+\frac{1}{{\tan }^2\left(\frac{3\pi }{16}\right)}$$$$\mathrm{let}\tan \left(\frac{\pi }{16}\right)=\mathrm{x}\mathrm{we}\mathrm{have}\mathrm{tanx}=\frac{2\tan \left(\frac{\mathrm{x}}{2}\right)}{1-{\tan }^2\left(\frac{\mathrm{x}}{2}\right)}\Rightarrow \mathrm{for}\mathrm{x}=\frac{\pi }{8}\mathrm{we}\mathrm{get}$$$$\sqrt{2}-1=\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\Rightarrow (\sqrt{2}-1)(1-{\mathrm{x}}^2)-2\mathrm{x}=0\Rightarrow$$$$(\sqrt{2}-1)+(1-\sqrt{2}){\mathrm{x}}^2-2\mathrm{x}=0\Rightarrow (1-\sqrt{2}){\mathrm{x}}^2-2\mathrm{x}+\sqrt{2}-1=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-(1-\sqrt{2})(\sqrt{2}-1)=1+{(\sqrt{2}-1)}^2=1+3-2\sqrt{2}=4-2\sqrt{2}$$$${\mathrm{x}}_1=\frac{1+\sqrt{4-2\sqrt{2}}}{1-\sqrt{2}}<0\mathrm{and}{\mathrm{x}}_2=\frac{1-\sqrt{4-2\sqrt{2}}}{1-\sqrt{2}}=\frac{-1+\sqrt{4-2\sqrt{2}}}{\sqrt{2}-1}>0\Rightarrow$$$$\tan \left(\frac{\pi }{16}\right)=\frac{\sqrt{4-2\sqrt{2}}-1}{\sqrt{2}-1}=(\sqrt{2}+1)(\sqrt{4-2\sqrt{2}}-1)\Rightarrow$$$${\tan }^2\left(\frac{\pi }{16}\right)=(3+2\sqrt{2})(4-2\sqrt{2}+1-2\sqrt{4-2\sqrt{2}})$$$$=(3+2\sqrt{2})(5-2\sqrt{2}-2\sqrt{4-2\sqrt{2}})\mathrm{rest}\mathrm{to}\mathrm{find}\tan \left(\frac{3\pi }{16}\right)...\mathrm{be}\mathrm{continued}...$$

Commented by Arijit last updated on 14/Feb/21

$$Thankyouverymuch$$

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