Question

1. \\(\frac{\tan^2 \theta}{1 + \tan^2 \theta}\\)

Explanation:

Step1: Recall trigonometric identity

We know the Pythagorean identity \(1 + \tan^{2}\theta=\sec^{2}\theta\), and also \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), \(\sec\theta=\frac{1}{\cos\theta}\). Substitute \(1 + \tan^{2}\theta=\sec^{2}\theta\) into the expression:
\(\frac{\tan^{2}\theta}{1 + \tan^{2}\theta}=\frac{\tan^{2}\theta}{\sec^{2}\theta}\)

Step2: Substitute \(\tan\theta\) and \(\sec\theta\) definitions

Since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and \(\sec\theta = \frac{1}{\cos\theta}\), then \(\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}\) and \(\sec^{2}\theta=\frac{1}{\cos^{2}\theta}\). Substitute these into the fraction:
\(\frac{\frac{\sin^{2}\theta}{\cos^{2}\theta}}{\frac{1}{\cos^{2}\theta}}\)

Step3: Simplify the fraction

When dividing by a fraction, we multiply by its reciprocal. So \(\frac{\frac{\sin^{2}\theta}{\cos^{2}\theta}}{\frac{1}{\cos^{2}\theta}}=\frac{\sin^{2}\theta}{\cos^{2}\theta}\times\cos^{2}\theta=\sin^{2}\theta\)

Answer:

\(\sin^{2}\theta\)