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If $\sin 2 \mathrm{~A}=\frac{4}{5}$, then what is the value of tan $\mathrm{A}\left(0 \leq \mathrm{A} \leq \frac{\pi}{4}\right)$ ?
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The correct answer is:
$\frac{1}{2}$
As given : $\sin 2 \mathrm{~A}=\frac{4}{5}$
$\sin 2 \mathrm{~A}=\frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{~A}}$
$\Rightarrow \frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{~A}}=\frac{4}{5}$
$\Rightarrow 10 \tan \mathrm{A}=4+4 \tan ^{2} \mathrm{~A}$
$\Rightarrow 5 \tan \mathrm{A}=2+2 \tan ^{2} \mathrm{~A}$
$\Rightarrow 2 \tan ^{2} \mathrm{~A}-5 \tan \mathrm{A}+2=0$
$\Rightarrow 2 \tan ^{2} \mathrm{~A}-4 \tan \mathrm{A}-\tan \mathrm{A}+2=0$
$\Rightarrow 2 \tan \mathrm{A}(\tan \mathrm{A}-2)-1(\tan \mathrm{A}-2)=0$
$\Rightarrow(2 \tan \mathrm{A}-1)(\tan \mathrm{A}-2)=0$
$\Rightarrow \tan \mathrm{A}=\frac{1}{2}$ (since $\left.\mathrm{A} \leq \frac{\pi}{4} \Rightarrow \tan \mathrm{A} \neq 2\right)$
$\sin 2 \mathrm{~A}=\frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{~A}}$
$\Rightarrow \frac{2 \tan \mathrm{A}}{1+\tan ^{2} \mathrm{~A}}=\frac{4}{5}$
$\Rightarrow 10 \tan \mathrm{A}=4+4 \tan ^{2} \mathrm{~A}$
$\Rightarrow 5 \tan \mathrm{A}=2+2 \tan ^{2} \mathrm{~A}$
$\Rightarrow 2 \tan ^{2} \mathrm{~A}-5 \tan \mathrm{A}+2=0$
$\Rightarrow 2 \tan ^{2} \mathrm{~A}-4 \tan \mathrm{A}-\tan \mathrm{A}+2=0$
$\Rightarrow 2 \tan \mathrm{A}(\tan \mathrm{A}-2)-1(\tan \mathrm{A}-2)=0$
$\Rightarrow(2 \tan \mathrm{A}-1)(\tan \mathrm{A}-2)=0$
$\Rightarrow \tan \mathrm{A}=\frac{1}{2}$ (since $\left.\mathrm{A} \leq \frac{\pi}{4} \Rightarrow \tan \mathrm{A} \neq 2\right)$
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