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

To prove the equation \( \frac{1 - \cos 2A}{\sin 2A} = \tan A \), we can start by using some trigonometric identities. First, recall the double angle formulas: 1. \( \cos 2A = 1 - 2\sin^2 A \) 2. \( \sin 2A = 2\sin A \cos A \) Now, substitute the formula for \( \cos 2A \) into the left-hand side of the equation: \[ 1 - \cos 2A = 1 - (1 - 2\sin^2 A) = 2\sin^2 A \] So now, the left-hand side becomes: \[ \frac{1 - \cos 2A}{\sin 2A} = \frac{2\sin^2 A}{\sin 2A} \] Next, we can substitute \( \sin 2A \) with \( 2\sin A \cos A \): \[ \frac{2\sin^2 A}{\sin 2A} = \frac{2\sin^2 A}{2\sin A \cos A} \] The \( 2 \) in the numerator and denominator cancels out: \[ \frac{2\sin^2 A}{2\sin A \cos A} = \frac{\sin^2 A}{\sin A \cos A} \] Now, we can simplify this further: \[ \frac{\sin^2 A}{\sin A \cos A} = \frac{\sin A}{\cos A} \] Recognizing this fraction gives us: \[ \frac{\sin A}{\cos A} = \tan A \] Thus, we conclude that: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] This completes the proof.