Q.If $\sin\theta = \dfrac{5}{13}$ and $\theta$ is an acute angle, find $\cos\theta$ and $\sec\theta$.

The Pythagorean identity gives $\cos\theta$ directly once $\sin\theta$ is known, and $\sec\theta$ then follows by taking a reciprocal.

$\cos^2\theta=1-\frac{25}{169}=\frac{144}{169}$, so $\cos\theta=\frac{12}{13}$ (positive, $\theta$ acute), and $\sec\theta=\frac{13}{12}$.

Step 1 — apply the Pythagorean identity.

Step 2 — take the square root, choosing the positive sign since $\theta$ is acute.

Step 3 — take the reciprocal for $\sec\theta$.

Equivalently, treat $\sin\theta=\frac{5}{13}$ as a right triangle with opposite = 5, hypotenuse = 13; then adjacent $=\sqrt{13^2-5^2}=12$ directly, giving the same values without naming the identity.

As with any square-root step, forgetting to check the sign against the given quadrant (here, acute) risks writing a negative value where $\theta$ is actually acute and both $\cos\theta$ and $\sec\theta$ must be positive.