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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