2 Prove the following trig identity \( \csc ^{2} \theta \tan ^{2} \theta-1=\tan ^{2} \theta \)

To prove the trigonometric identity: \[ \csc^{2} \theta \tan^{2} \theta - 1 = \tan^{2} \theta \] we will start by expressing all trigonometric functions in terms of sine and cosine. This approach often simplifies the verification of trigonometric identities. ### Step 1: Express \(\csc \theta\) and \(\tan \theta\) in Terms of \(\sin \theta\) and \(\cos \theta\) Recall the following definitions: \[ \csc \theta = \frac{1}{\sin \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \] Squaring both expressions: \[ \csc^{2} \theta = \left(\frac{1}{\sin \theta}\right)^2 = \frac{1}{\sin^{2} \theta} \] \[ \tan^{2} \theta = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\sin^{2} \theta}{\cos^{2} \theta} \] ### Step 2: Substitute Back into the Original Identity Substitute \(\csc^{2} \theta\) and \(\tan^{2} \theta\) into the left-hand side (LHS) of the identity: \[ \csc^{2} \theta \tan^{2} \theta - 1 = \left(\frac{1}{\sin^{2} \theta}\right) \left(\frac{\sin^{2} \theta}{\cos^{2} \theta}\right) - 1 \] Simplify the multiplication: \[ \frac{1}{\sin^{2} \theta} \cdot \frac{\sin^{2} \theta}{\cos^{2} \theta} = \frac{\sin^{2} \theta}{\sin^{2} \theta \cos^{2} \theta} = \frac{1}{\cos^{2} \theta} = \sec^{2} \theta \] So, the expression becomes: \[ \sec^{2} \theta - 1 \] ### Step 3: Apply the Pythagorean Identity We know the Pythagorean identity: \[ \sec^{2} \theta - 1 = \tan^{2} \theta \] Therefore: \[ \csc^{2} \theta \tan^{2} \theta - 1 = \tan^{2} \theta \] ### Conclusion Through substitution and the application of fundamental trigonometric identities, we have successfully proven that: \[ \csc^{2} \theta \tan^{2} \theta - 1 = \tan^{2} \theta \]