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Question Number 164391 by bobhans last updated on 16/Jan/22

$$\mathrm{If}\{\begin{array}{l}\sin (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{10}}\\ \cos (\mathrm{A}+\mathrm{B})=\frac{2}{\sqrt{29}}\end{array};0<\mathrm{A}<\frac{\pi }{4};0<\mathrm{B}<\frac{\pi }{4}$$ $$\mathrm{Find}\tan 2\mathrm{A}.$$

Answered by cortano1 last updated on 16/Jan/22

$$\begin{array}{c}\sin (A-B)=\frac{1}{\sqrt{10}}\\ \cos (A+B)=\frac{2}{\sqrt{29}}\end{array}\}\Rightarrow \{\begin{array}{l}0<A+B<\frac{\pi }{2}\\ -\frac{\pi }{4}<A-B<\frac{\pi }{4}\end{array}$$ $$\{\begin{array}{l}\tan (A-B)=\frac{1}{3}\\ \tan (A+B)=\frac{5}{2}\end{array}$$ $$\Rightarrow \tan 2A=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B).\tan (A-B)}$$ $$\tan 2A=\frac{\frac{5}{2}+\frac{1}{3}}{1-\frac{5}{6}}=17$$

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