Without the use of tables or a calculator prove that \( \frac{1-\cos 2 A}{\sin 2 A}=\tan A \)

To prove the identity: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] **We will use fundamental trigonometric identities to simplify the left-hand side (LHS) and show that it equals the right-hand side (RHS).** ### Step 1: Recall the Double-Angle Identities The double-angle identities for cosine and sine are: 1. **Cosine Double-Angle Identity:** \[ \cos 2A = 1 - 2\sin^2 A \quad \text{or} \quad \cos 2A = 2\cos^2 A - 1 \] 2. **Sine Double-Angle Identity:** \[ \sin 2A = 2\sin A \cos A \] ### Step 2: Express \(1 - \cos 2A\) in Terms of \(\sin A\) Using the first form of the cosine double-angle identity: \[ \cos 2A = 1 - 2\sin^2 A \\ \Rightarrow 1 - \cos 2A = 2\sin^2 A \] So, \[ 1 - \cos 2A = 2\sin^2 A \] ### Step 3: Substitute the Expressions into the LHS Now, substitute \(1 - \cos 2A\) and \(\sin 2A\) into the original expression: \[ \frac{1 - \cos 2A}{\sin 2A} = \frac{2\sin^2 A}{2\sin A \cos A} \] ### Step 4: Simplify the Expression Cancel the common factor of 2 in the numerator and the denominator: \[ \frac{2\sin^2 A}{2\sin A \cos A} = \frac{\sin^2 A}{\sin A \cos A} \] Simplify \(\sin^2 A\) divided by \(\sin A\): \[ \frac{\sin^2 A}{\sin A \cos A} = \frac{\sin A}{\cos A} \] ### Step 5: Recognize the Tangent Function The ratio \(\frac{\sin A}{\cos A}\) is the definition of the tangent function: \[ \frac{\sin A}{\cos A} = \tan A \] ### Conclusion Thus, we have shown that: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] **Q.E.D.**