Given-A-B-arctan-1-2-A-B-arctan-1-3-then-tan-A-

Question Number 128641 by john_santu last updated on 09/Jan/21

Given {\begin{cases} A + B = \arctan \frac{1}{2} \\ A - B = \arctan \frac{1}{3} \end{cases}

then \tan A = ?

Answered by liberty last updated on 09/Jan/21

(1) + (2) \Rightarrow 2 A = \arctan \frac{1}{2} + \arctan \frac{1}{3}

\tan 2 A = \frac{1 / 2 + 1 / 3}{1 - (1 / 2) \times (1 / 3)} = 1

\Rightarrow \frac{2 \tan A}{1 - \tan^{2} A} = 1; \tan^{2} A + 2 \tan A = 1

{(\tan A + 1)}^{2} = 2; \tan A = - 1 + \sqrt{2}

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