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Question Number 113769 by deleteduser12 last updated on 15/Sep/20

$$provethat,\tan \left(7\frac{1}{2}\right)°=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

$$tan\left(\frac{\pi }{24}\right)=\frac{sin\frac{\pi }{24}}{cos\frac{\pi }{24}}=\frac{2{sin}^2\frac{\pi }{24}}{2sin\frac{\pi }{24}cos\frac{\pi }{24}}=\frac{1-cos\frac{\pi }{12}}{sin\frac{\pi }{12}}$$$$cos\frac{\pi }{12}=cos(\frac{\pi }{3}-\frac{\pi }{4})=\frac{\sqrt{3}+1}{2\sqrt{2}}$$$$sin\frac{\pi }{12}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$$$So\frac{1-cos\frac{\pi }{12}}{sin\frac{\pi }{12}}=\frac{2\sqrt{2}-\sqrt{3}+1}{\sqrt{3}-1}=\frac{2\sqrt{2}(\sqrt{3}+1)}{2}-(2+\sqrt{3})=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$$$$$$$$$$

Answered by $@y@m last updated on 15/Sep/20

$$\tan {30}^{\mathrm{o}}=\frac{2\tan {15}^{\mathrm{o}}}{1-\tan {}^2{15}^{\mathrm{o}}}$$$$\frac{1}{\sqrt{3}}=\frac{2x}{1-x^2}wherex=\tan {15}^o$$$$1-x^2=2\sqrt{3}x$$$$x^2+2\sqrt{3}x-1=0$$$$x=\frac{-2\sqrt{3}\pm \sqrt{12+4}}{2}$$$$x=\frac{-2\sqrt{3}\pm 4}{2}=-\sqrt{3}\pm 2$$$$But\tan {15}^o>0$$$$\therefore \tan {15}^o=2-\sqrt{3}...\left(1\right)$$$$Similarly,$$$$\tan {15}^{\mathrm{o}}=\frac{2\tan {\left(7\frac{1}{2}\right)}^{\mathrm{o}}}{1-\tan {}^2{\left(7\frac{1}{2}\right)}^{\mathrm{o}}}$$$$2-\sqrt{3}=\frac{2y}{1-y^2}wherey=\tan {\left(7\frac{1}{2}\right)}^{\mathrm{o}}$$$$(2-\sqrt{3})y^2+2y-(2-\sqrt{3})=0$$$$y=\frac{-2\pm \sqrt{4+4(4+3-4\sqrt{3})}}{2(2+\sqrt{3})}$$$$y=\frac{-2\pm 2\sqrt{1+(7-4\sqrt{3})}}{2(2-\sqrt{3})}$$$$y=\frac{-1\pm \sqrt{8-4\sqrt{3}}}{2+\sqrt{3}}$$$$y=\frac{-1\pm (\sqrt{6}-\sqrt{2})}{2+\sqrt{3}}$$$$Buty>0$$$$\therefore y=\frac{\sqrt{6}-\sqrt{2}-1}{2-\sqrt{3}}$$$$y=\frac{\sqrt{6}-\sqrt{2}-1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}$$$$y=\frac{2\sqrt{6}-2\sqrt{2}-2-\sqrt{6}+\sqrt{18}-\sqrt{3}}{2^2-\sqrt{3^2}}$$$$y=\sqrt{6}-2\sqrt{2}-2+3\sqrt{2}-\sqrt{3}$$$$y=\sqrt{6}+\sqrt{2}-\sqrt{3}-2$$

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