Question Number 164391 by bobhans last updated on 16/Jan/22
$$\:\:\mathrm{If}\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}}\end{cases};\:\mathrm{0}<\mathrm{A}<\frac{\pi}{\mathrm{4}}\:;\:\mathrm{0}<\mathrm{B}<\frac{\pi}{\mathrm{4}} \\ $$$$\:\mathrm{Find}\:\mathrm{tan}\:\mathrm{2A}. \\ $$
Answered by cortano1 last updated on 16/Jan/22
$$\:\:\left.\begin{matrix}{\mathrm{sin}\:\left({A}−{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{cos}\:\left({A}+{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}}\end{matrix}\right\}\:\Rightarrow\begin{cases}{\mathrm{0}<{A}+{B}<\frac{\pi}{\mathrm{2}}}\\{−\frac{\pi}{\mathrm{4}}<{A}−{B}<\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{tan}\:\left({A}−{B}\right)=\frac{\mathrm{1}}{\mathrm{3}}}\\{\mathrm{tan}\:\left({A}+{B}\right)=\frac{\mathrm{5}}{\mathrm{2}}}\end{cases} \\ $$$$\:\Rightarrow\mathrm{tan}\:\mathrm{2}{A}=\:\frac{\mathrm{tan}\:\left({A}+{B}\right)+\mathrm{tan}\:\left({A}−{B}\right)}{\mathrm{1}−\mathrm{tan}\:\left({A}+{B}\right).\mathrm{tan}\:\left({A}−{B}\right)} \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{2}{A}\:=\:\frac{\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{5}}{\mathrm{6}}}\:=\:\mathrm{17} \\ $$