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Question Number 154240 by amin96 last updated on 15/Sep/21

$$\{\begin{array}{l}tg(a+b)=7\\ tg(a-b)=5\end{array}tg\left(a\right)=?easyquestion$$

Answered by mr W last updated on 15/Sep/21

$$a+b={\tan }^{-1}7$$$$a-b={\tan }^{-1}5$$$$2a={\tan }^{-1}7+{\tan }^{-1}5$$$$\tan 2a=\frac{7+5}{1-7\times 5}=-\frac{6}{17}=\frac{2\tan a}{1-{\tan }^{2}a}$$$${3\tan }^{2}a-17\tan a-3=0$$$$\Rightarrow \tan a=\frac{17\pm 5\sqrt{13}}{6}$$

Commented by SLVR last updated on 26/Sep/21

$$sir..{tan}^{-1}7+{tan}^{-1}5canbe$$$$\pi +{tan}^{-1}\left(\frac{7+5}{1-7\times 5}\right)..ihope$$

Commented by SLVR last updated on 26/Sep/21

$$itsaconditionx>0y>0and$$$$ifxy>1itcanbe\pi +{tan}^{-1}\left(\right).$$

Commented by mr W last updated on 26/Sep/21

$$that{doesn}^{\prime }taffecttheresult,$$$$sincewetake\tan (...)attheend.$$$$hereyoucansee:$$$$a+b=n\pi +{\tan }^{-1}7$$$$a-b=m\pi +{\tan }^{-1}5$$$$2a=n\pi +{\tan }^{-1}7+m\pi +{\tan }^{-1}5$$$$\tan 2a=\frac{\tan (n\pi +{\tan }^{-1}7)+\tan (m\pi +{\tan }^{-1}5)}{1-\tan (n\pi +{\tan }^{-1}7)\times \tan (m\pi +{\tan }^{-1}5)}$$$$\tan 2a=\frac{7+5}{1-7\times 5}=-\frac{6}{17}=\frac{2\tan a}{1-{\tan }^{2}a}$$

Commented by SLVR last updated on 26/Sep/21

$$wow...thatsright...prof...W$$$$$$

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