1.2 If \( \tan \theta=t \) and \( \theta \) is acute, determine \( \sin \theta \) in terms of \( t \)

To find \( \sin \theta \) in terms of \( t \) when \( \tan \theta = t \), we start from the definition of tangent. We know that: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] Given \( \tan \theta = t \), we can express it as: \[ \frac{\sin \theta}{\cos \theta} = t \] From this, we can express \( \sin \theta \): \[ \sin \theta = t \cos \theta \] Now, we also use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting \( \sin \theta = t \cos \theta \) into the identity gives: \[ (t \cos \theta)^2 + \cos^2 \theta = 1 \] This simplifies to: \[ t^2 \cos^2 \theta + \cos^2 \theta = 1 \] Factoring out \( \cos^2 \theta \): \[ \cos^2 \theta (t^2 + 1) = 1 \] Thus, we find: \[ \cos^2 \theta = \frac{1}{t^2 + 1} \] Taking the square root (and considering \( \theta \) is acute, hence \( \cos \theta > 0 \)): \[ \cos \theta = \frac{1}{\sqrt{t^2 + 1}} \] Now we substitute \( \cos \theta \) back into our expression for \( \sin \theta \): \[ \sin \theta = t \cos \theta = t \cdot \frac{1}{\sqrt{t^2 + 1}} \] Therefore, we conclude that: \[ \sin \theta = \frac{t}{\sqrt{t^2 + 1}} \]