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Prove \( \frac{1-\cos 2 A}{\sin 2 A}=\tan \)

Ask by Bowman Clark. in South Africa
Feb 02,2025

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Answer

\[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \]

Solution

Certainly! Let's prove the trigonometric identity: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] **Step 1: Recall Double Angle Identities** First, recall the double angle identities for sine and cosine: \[ \cos 2A = 1 - 2\sin^2 A \quad \text{and} \quad \sin 2A = 2\sin A \cos A \] **Step 2: Substitute the Double Angle Identity for \(\cos 2A\)** Substitute \(\cos 2A = 1 - 2\sin^2 A\) into the left-hand side (LHS) of the equation: \[ \frac{1 - \cos 2A}{\sin 2A} = \frac{1 - (1 - 2\sin^2 A)}{\sin 2A} = \frac{2\sin^2 A}{\sin 2A} \] **Step 3: Substitute the Double Angle Identity for \(\sin 2A\)** Now, substitute \(\sin 2A = 2\sin A \cos A\) into the equation: \[ \frac{2\sin^2 A}{2\sin A \cos A} = \frac{\sin^2 A}{\sin A \cos A} \] **Step 4: Simplify the Expression** Cancel out \(\sin A\) from the numerator and denominator: \[ \frac{\sin A}{\cos A} = \tan A \] **Conclusion** Thus, we have shown that: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] **Final Identity:** \[ \boxed{\frac{1 - \cos 2A}{\sin 2A} = \tan A} \]

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To prove the identity \( \frac{1 - \cos 2A}{\sin 2A} = \tan A \), let's start with some trigonometric identities. Recall that: 1. \( \cos 2A = 1 - 2\sin^2 A \) (the double angle formula for cosine). 2. \( \sin 2A = 2\sin A \cos A \) (the double angle formula for sine). Now substituting \( \cos 2A \) into the left side of the equation: \[ \frac{1 - \cos 2A}{\sin 2A} = \frac{1 - (1 - 2 \sin^2 A)}{2 \sin A \cos A} \] Simplifying the numerator: \[ = \frac{2 \sin^2 A}{2 \sin A \cos A} \] The \( 2 \) in both the numerator and denominator can be canceled: \[ = \frac{\sin^2 A}{\sin A \cos A} \] Now we can simplify this further: \[ = \frac{\sin A}{\cos A} \] This is equal to \( \tan A \). Hence, we have: \[ \frac{1 - \cos 2A}{\sin 2A} = \tan A \] This completes the proof.

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