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Question Number 131302 by EDWIN88 last updated on 03/Feb/21

$$Given\sin (\frac{\pi }{6}+{\tan }^{-1}\left(x\right))=\frac{13}{14}$$$$If{\tan }^{-1}\left(x\right)={\tan }^{-1}\left(\frac{a\sqrt{3}}{b}\right)then$$$$\frac{a+b}{2}=?$$

Answered by mr W last updated on 03/Feb/21

$$t={\tan }^{-1}x<\frac{\pi }{3}!$$$$\sin (\frac{\pi }{6}+t)=\frac{13}{14}$$$$\frac{1}{2}\cos t+\frac{\sqrt{3}}{2}\sin t=\frac{13}{14}$$$$\cos t+\sqrt{3}\sin t=\frac{13}{7}$$$$u=\cos t$$$$\sqrt{3(1-u^2)}=\frac{13}{7}-u$$$$3(1-u^2)=\frac{169}{49}+u^2-\frac{26}{7}u$$$$2u^2-\frac{13}{7}u+\frac{11}{49}=0$$$$u=\frac{1}{4}\left(\frac{13\pm 9}{7}\right)=\frac{11}{14},\frac{1}{7}$$$$\tan t=\sqrt{\frac{1}{{\cos }^{2}t}-1}=\frac{5\sqrt{3}}{11},4\sqrt{3}>\sqrt{3}!rejected$$$$\Rightarrow t={\tan }^{-1}\frac{5\sqrt{3}}{11}$$$${\tan }^{-1}x=t={\tan }^{-1}\frac{5\sqrt{3}}{11}$$$$\Rightarrow a=5,b=11$$$$\Rightarrow \frac{a+b}{2}=8$$

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