Find $\tan(\arctan(\frac{1}{3})+\arctan(\frac{1}{9}))$

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Why is this incorrect? Everything is according to the formula: $$\tan(\arctan(\frac{1}{3})+\arctan(\frac{1}{9}))=\frac{\tan(\arctan(\frac{1}{3}))+\tan(\arctan(\frac{1}{9}))}{1-\tan(\arctan(\frac{1}{3}))\tan(\frac{1}{9})}=\frac{1}{2}$$

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Evaluating the RHS is $$ \frac{1/3 + 1/9}{1 - 1/27} = \frac{27}{26} \frac{4}{9} = \frac{6}{13} \ne 1/2 $$