If frac{pi}{4} + sum_{p=1}^{11} tan^{-1} left | vedclass

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(A) We use the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$.Given the term $	an^{-1} \left( \frac{2^{p-1}}{1+2^{2

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$2048$

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(A) We use the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$.Given the term $	an^{-1} \left( \frac{2^{p-1}}{1+2^{2p-1}} ight)$, we can rewrite it as $	an^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} ight) = 	an^{-1}(2^p) - 	an^{-1}(2^{p-1})$.Now, sum the telescoping series: $\sum_{p=1}^{11} (	an^{-1}(2^p) - 	an^{-1}(2^{p-1})) = (	an^{-1}(2^1) - 	an^{-1}(2^0)) + (	an^{-1}(2^2) - 	an^{-1}(2^1)) + \dots + (	an^{-1}(2^{11}) - 	an^{-1}(2^{10}))$.This simplifies to $	an^{-1}(2^{11}) - 	an^{-1}(2^0) = 	an^{-1}(2048) - \frac{\pi}{4}$.Substituting this back into the original equation: $\frac{\pi}{4} + (	an^{-1}(2048) - \frac{\pi}{4}) = 	an^{-1}(2048)$.Thus, $	an^{-1} \alpha = 	an^{-1}(2048)$, which implies $\alpha = 2048$.Therefore, $	an \alpha = 	an(2048)$.

If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left(\frac{2^{p-1}}{1+2^{2p-1}}\right) = \tan^{-1} \alpha$, then $\tan \alpha$ is equal to . . . . . . .

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