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Question Number 83680 by Power last updated on 05/Mar/20

Commented by john santu last updated on 05/Mar/20

$$\mathrm{considering}\mathrm{A}+\mathrm{B}={30}^{\mathrm{o}}$$$$\mathrm{B}={30}^{\mathrm{o}}-\mathrm{A}\Rightarrow \tan \mathrm{B}=\frac{\frac{1}{\sqrt{3}}-\tan \mathrm{A}}{1+\frac{1}{\sqrt{3}}\tan \mathrm{A}}$$$$\tan \mathrm{B}=\frac{1-\sqrt{3}\tan \mathrm{A}}{\sqrt{3}+\tan \mathrm{A}}$$$$\sqrt{3}+\tan \mathrm{B}=\sqrt{3}+\frac{1-\sqrt{3}\tan \mathrm{A}}{\sqrt{3}+\tan \mathrm{A}}$$$$=\frac{4}{\sqrt{3}+\tan \mathrm{A}}$$$$\mathrm{now}(\sqrt{3}+\tan \mathrm{A})(\sqrt{3}+\tan \mathrm{B})=$$$$(\sqrt{3}+\tan \mathrm{A})\left(\frac{4}{\sqrt{3}+\tan \mathrm{A}}\right)=4$$$$\mathrm{so}\Rightarrow \left(4^{14}\right)(\sqrt{3}+\tan {15}^{\mathrm{o}})$$$$\mathrm{i}\mathrm{think}\mathrm{it}\mathrm{easy}$$$$$$

Commented by jagoll last updated on 05/Mar/20

$$\tan {15}^{\mathrm{o}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{4-2\sqrt{3}}{2}$$$$=2-\sqrt{3}$$$$\mathrm{finally}=4^{14}(\sqrt{3}+2-\sqrt{3})=4^{14}\times 2$$$$\mathrm{sorry}\mathrm{mistake}$$

Commented by MJS last updated on 05/Mar/20

$$\mathrm{not}14\times 4\mathrm{but}{4}^{14}$$$$\Rightarrow \mathrm{answer}\mathrm{is}2\times 4^{14}=2^{29}=536870912$$

Commented by john santu last updated on 05/Mar/20

$$\mathrm{oo}\mathrm{yes}\mathrm{sir}.4\times 4\times 4\times ...\times 4\left(14\mathrm{times}\right)$$

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