If tan^{-1}(1) + tan^{-1}(3) + tan^{-1}(5) + | vedclass

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(A) Step $1$: Use the property $	an^{-1}(x) + 	an^{-1}(y) = \pi + 	an^{-1}(\frac{x+y}{1-xy})$ for $xy > 1$.Step $2$: Combine $	an^{-1}(3) + 	an^{

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$46$

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(A) Step $1$: Use the property $	an^{-1}(x) + 	an^{-1}(y) = \pi + 	an^{-1}(\frac{x+y}{1-xy})$ for $xy > 1$.Step $2$: Combine $	an^{-1}(3) + 	an^{-1}(5) = \pi + 	an^{-1}(\frac{3+5}{1-15}) = \pi + 	an^{-1}(\frac{8}{-14}) = \pi - 	an^{-1}(4/7)$.Step $3$: The expression becomes $	an^{-1}(1) + \pi - 	an^{-1}(4/7) + 	an^{-1}(1/4) = \pi + 	an^{-1}(1) + 	an^{-1}(1/4) - 	an^{-1}(4/7)$.Step $4$: Use $	an^{-1}(1) + 	an^{-1}(1/4) = 	an^{-1}(\frac{1+1/4}{1-1/4}) = 	an^{-1}(\frac{5/4}{3/4}) = 	an^{-1}(5/3)$.Step $5$: Now, $	an^{-1}(5/3) - 	an^{-1}(4/7) = 	an^{-1}(\frac{5/3 - 4/7}{1 + (5/3)(4/7)}) = 	an^{-1}(\frac{35-12}{21+20}) = 	an^{-1}(23/41)$.Step $6$: Comparing with $	an^{-1}(\alpha/2)$, we have $\alpha/2 = 23/41$, so $\alpha = 46/41$.

If $\tan^{-1}(1) + \tan^{-1}(3) + \tan^{-1}(5) + \tan^{-1}(1/4) = \pi + \tan^{-1}(\alpha/2)$, then the value of $\alpha$ is... (in $/41$)

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