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Question Number 227232 by fantastic2 last updated on 07/Jan/26

(64−(8π+4π))−(8π−(4(tan^(−1) 2+4tan^(−1) (1/2)−2)+8(π−2)))−(64−(16π+8π−16(tan^(−1) 2+4tan^(−1) (1/2)−2)))−(3.2+2+(32tan^(−1) ((4.8)/(6.4))−32sin tan^(−1) ((4.8)/(6.4)))−π)+4(tan^(−1) 2+4tan^(−1) (1/2)−2)

$$\left(\mathrm{64}−\left(\mathrm{8}\pi+\mathrm{4}\pi\right)\right)−\left(\mathrm{8}\pi−\left(\mathrm{4}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{4tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)+\mathrm{8}\left(\pi−\mathrm{2}\right)\right)\right)−\left(\mathrm{64}−\left(\mathrm{16}\pi+\mathrm{8}\pi−\mathrm{16}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{4tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)\right)\right)−\left(\mathrm{3}.\mathrm{2}+\mathrm{2}+\left(\mathrm{32tan}^{−\mathrm{1}} \frac{\mathrm{4}.\mathrm{8}}{\mathrm{6}.\mathrm{4}}−\mathrm{32sin}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}.\mathrm{8}}{\mathrm{6}.\mathrm{4}}\right)−\pi\right)+\mathrm{4}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{4tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right) \\ $$

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