Q.If $\cos\theta=\dfrac{5}{13}$ and $\theta$ is acute, find the value of $\tan2\theta$.

With $\theta$ acute, the $5$-$12$-$13$ Pythagorean triple gives $\sin\theta$ and hence $\tan\theta$ directly, which then substitutes into the double-angle tangent formula.

Step 1 — find $\sin\theta$ and $\tan\theta$. $\sin^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}$, so $\sin\theta=\dfrac{12}{13}$ (positive, $\theta$ acute) — the familiar $5$-$12$-$13$ Pythagorean triple. Then $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{12/13}{5/13}=\dfrac{12}{5}$.

Step 2 — apply the double-angle tangent formula. $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2\cdot\frac{12}{5}}{1-\left(\frac{12}{5}\right)^2}=\dfrac{\frac{24}{5}}{1-\frac{144}{25}}=\dfrac{\frac{24}{5}}{\frac{25-144}{25}}=\dfrac{\frac{24}{5}}{-\frac{119}{25}}$.

Step 3 — simplify. $=\dfrac{24}{5}\times\left(-\dfrac{25}{119}\right)=-\dfrac{24\times5}{119}=-\dfrac{120}{119}$. …

Sign errors when squaring $\tan\theta=\dfrac{12}{5}$ and simplifying $1-\dfrac{144}{25}$ (getting the order of subtraction backwards); also forgetting that $2\theta$ can land in a different quadrant fro …