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

To prove the trigonometric identity: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] **without using tables or a calculator**, we'll utilize fundamental trigonometric identities. Here's a step-by-step demonstration: ### **Step 1: Recall the Double-Angle Formulas** The double-angle formulas for sine and cosine are essential for this proof: 1. **Sine Double-Angle Identity**: \[ \sin 2A = 2 \sin A \cos A \] 2. **Cosine Double-Angle Identity**: \[ \cos 2A = 1 - 2 \sin^2 A \] *(Alternatively, \(\cos 2A = 2 \cos^2 A - 1\) can also be used, but we'll proceed with the first form.)* ### **Step 2: Substitute the Cosine Double-Angle Identity into the Numerator** Starting with the numerator of the left-hand side (LHS): \[ 1 - \cos 2A = 1 - (1 - 2 \sin^2 A) = 2 \sin^2 A \] So, the expression becomes: \[ \frac{1 - \cos 2A}{\sin 2A} = \frac{2 \sin^2 A}{\sin 2A} \] ### **Step 3: Substitute the Sine Double-Angle Identity into the Denominator** Replace \(\sin 2A\) with its double-angle equivalent: \[ \frac{2 \sin^2 A}{\sin 2A} = \frac{2 \sin^2 A}{2 \sin A \cos A} \] ### **Step 4: Simplify the Fraction** Cancel out the common factors in the numerator and the denominator: \[ \frac{2 \sin^2 A}{2 \sin A \cos A} = \frac{\sin^2 A}{\sin A \cos A} = \frac{\sin A}{\cos A} \] *(Here, \(2\) in the numerator and denominator cancels out, and one \(\sin A\) in the numerator cancels with one \(\sin A\) in the denominator.)* ### **Step 5: Recognize the Tangent Function** The simplified expression is the definition of the tangent function: \[ \frac{\sin A}{\cos A} = \tan A \] ### **Conclusion** Thus, we've shown that: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] This completes the proof.