Question
If $\cos\text{x}=\frac{4}{5}$ and x is acute, find $\tan 2\text{x}$

Answer

$\text{since}\ \theta\ \text{in acute},\text{so}\ 0\leq2\theta<\pi$
Now,
$\text{cos}\theta=\frac{4}{5}=\frac{\text{b}}{\text{h}}\Rightarrow\text{p}=3$
$\text{h}=5$
$\therefore\sin\theta=\frac{\text{p}}{\text{h}}=\frac{3}{5}$
$\tan\theta=\frac{\text{p}}{\text{b}}=\frac{3}{4}$
so,
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{2.\frac{3}{4}}{1-\Big(\frac{3}{4}\Big)^2}$
$=\frac{\frac{6}{4}}{\frac{7}{16}}=\frac{24}{7}$

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