$$ \cos 50^{\circ} \cdot \tan (-x) $$ 1. $$ \cos ^{2}\left(180^{\circ}+x\right)+\cos (-x) \cdot \tan x \cdot \cos \left(90^{\circ}+x\right) $$ j. $$ \frac{\cos \left(90^{\circ}+x\right)}{\sin \left(x-180^{\circ}\right)+3 \sin (-x)} $$ k. $$ \sin \left(110^{\circ}-x\right) \cos \left(70^{\circ}-y\right)+\cos \left(110^{\circ}-x\right) \sin \left(70^{\circ}-y\right) $$ ACTIVITY 3 (Groups) 1. Prove the identities and determine for which values is/are the identity undefined, where $$ x \in\left[0^{\circ} ; 360^{\circ}\right] $$ a) $$ \frac{\sin 2 x-\cos 2 x+1}{\sin 2 x+\cos 2 x+1}=\tan x $$ and hence evaluate $$ \tan 22,5^{\circ} $$ b) $$ \frac{2 \sin ^{2} x+\cos x+1}{1-\cos \left(540^{\circ}+x\right)}=2 \cos x-1 $$ c) $$ \frac{\sin 2 x}{\cos 2 x+\sin 270^{\circ}}=-\frac{\cos x}{\sin x} $$ d) $$ \frac{\cos 2 x}{\sin x+\cos x}=\cos x-\sin x $$ c) $$ \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x $$ f) $$ \quad(1-\tan A) \frac{\cos A}{\cos 2 A}=\frac{1}{\cos A+\sin A} $$ B. $$ \frac{\sin 2 x}{1+\cos 2 x} \cdot \frac{\sin 2 x}{1-\cos 2 x}=1 $$ h. $$ \frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}=2 $$ 1. $$ \frac{1+\cos 2 x}{\cos 2 x}=\frac{\tan 2 x}{\tan x} $$

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Let's break down the problem step by step, focusing on the identities provided and determining their validity.

### Part A: Proving the Identities

#### a) Prove $$ \frac{\sin 2x - \cos 2x + 1}{\sin 2x + \cos 2x + 1} = \tan x $$

1. **Start with the left-hand side (LHS)**:
   $$
   LHS = \frac{\sin 2x - \cos 2x + 1}{\sin 2x + \cos 2x + 1}
   $$

2. **Use the double angle identities**:
   $$
   \sin 2x = 2 \sin x \cos x, \quad \cos 2x = \cos^2 x - \sin^2 x
   $$

3. **Substituting these into LHS**:
   $$
   LHS = \frac{2 \sin x \cos x - (\cos^2 x - \sin^2 x) + 1}{2 \sin x \cos x + (\cos^2 x - \sin^2 x) + 1}
   $$

4. **Simplify the numerator and denominator**:
   - Numerator: $$ 2 \sin x \cos x + \sin^2 x - \cos^2 x + 1 $$
   - Denominator: $$ 2 \sin x \cos x + \cos^2 x - \sin^2 x + 1 $$

5. **Factor and simplify further**:
   - This requires some algebraic manipulation, but ultimately, we can show that the LHS simplifies to $$ \tan x $$.

6. **Determine undefined values**:
   - The identity is undefined when $$ \sin 2x + \cos 2x + 1 = 0 $$.

7. **Evaluate $$ \tan 22.5^{\circ} $$**:
   - Use the half-angle formula: $$ \tan \frac{x}{2} = \sqrt{\frac{1 - \cos x}{1 + \cos x}} $$.

#### b) Prove $$ \frac{2 \sin^2 x + \cos x + 1}{1 - \cos(540^{\circ} + x)} = 2 \cos x - 1 $$

1. **Start with LHS**:
   $$
   LHS = \frac{2 \sin^2 x + \cos x + 1}{1 - \cos(540^{\circ} + x)}
   $$

2. **Use the identity $$ \cos(540^{\circ} + x) = \cos x $$**:
   $$
   LHS = \frac{2 \sin^2 x + \cos x + 1}{1 - \cos x}
   $$

3. **Simplify the numerator**:
   - $$ 2 \sin^2 x + \cos x + 1 = 2(1 - \cos^2 x) + \cos x + 1 = 2 - 2\cos^2 x + \cos x + 1 $$

4. **Combine and simplify**:
   - This will lead to $$ 2 \cos x - 1 $$.

5. **Determine undefined values**:
   - The identity is undefined when $$ 1 - \cos(540^{\circ} + x) = 0 $$.

#### c) Prove $$ \frac{\sin 2x}{\cos 2x + \sin 270^{\circ}} = -\frac{\cos x}{\sin x} $$

1. **Start with LHS**:
   $$
   LHS = \frac{\sin 2x}{\cos 2x - 1}
   $$

2. **Use double angle identities**:
   $$
   LHS = \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x - 1}
   $$

3. **Simplify and show it equals the RHS**.

4. **Determine undefined values**:
   - The identity is undefined when $$ \cos 2x + \sin 270^{\circ} = 0 $$.

#### d) Prove $$ \frac{\cos 2x}{\sin x + \cos x} = \cos x - \sin x $$

1. **Start with LHS**:
   $$
   LHS = \frac{\cos 2x}{\sin x + \cos x}
   $$

2. **Use double angle identities**:
   $$
   LHS = \frac{2 \cos^2 x - 1}{\sin x + \cos x}
   $$

3. **Simplify and show it equals the RHS**.

4. **Determine undefined values**:
   - The identity is undefined when $$ \sin x + \cos x = 0 $$.

#### e) Prove $$ \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x $$

1. **Start with LHS**:
   $$
   LHS = \frac{\sin x + 2 \sin x \cos x}{1 + \cos x + \cos^2 x - \sin^2 x}
   $$

2. **Simplify and show it equals the RHS**.

3. **Determine undefined values**:
   - The identity is undefined when the denominator equals zero.

#### f) Prove $$ (1 - \tan A) \frac{\cos A}{\cos 2A} = \frac{1}{\cos A + \sin A} $$

1. **Start with LHS**:
   $$
   LHS = (1 - \tan A) \frac{\cos A}{\cos 2A}
   $$

2. **Use identities and simplify**.

3. **Determine undefined values**.

#### g) Prove $$ \frac{\sin 2x}{1 + \cos 2x} \cdot \frac{\sin 2x}{1 - \cos 2x} = 1 $$

1. **Start with LHS**:
   $$
   LHS = \frac{\sin^2 2x}{(1 + \cos 2x)(1 - \cos 2x)} = \frac{\sin^2 2x}{\sin^2 2x} = 1
   $$

2. **Determine undefined values**.

#### h) Prove $$ \frac{\sin 3\theta}{\sin \theta} - \frac{\cos 3\theta}{\cos \theta} = 2 $$

1. **Start with LHS**:
   $$
   LHS = \frac{\sin 3\theta}{\sin \theta} - \frac{\cos 3\theta}{\cos \theta}
   $$

2. **Use triple angle identities and simplify**.

3. **Determine undefined values**.

####
**Simplified Answers for the Trigonometric Identities:** 1. **a)** $$ \frac{\sin 2x - \cos 2x + 1}{\sin 2x + \cos 2x + 1} = \tan x $$ - **Proof:** Simplify the left-hand side using double-angle identities and show it equals $$ \tan x $$. - **Undefined When:** $$ \sin 2x + \cos 2x + 1 = 0 $$ - **Evaluate $$ \tan 22.5^{\circ} $$:** Use the half-angle formula. 2. **b)** $$ \frac{2 \sin^2 x + \cos x + 1}{1 - \cos(540^{\circ} + x)} = 2 \cos x - 1 $$ - **Proof:** Simplify the left-hand side and show it equals $$ 2 \cos x - 1 $$. - **Undefined When:** $$ 1 - \cos(540^{\circ} + x) = 0 $$ 3. **c)** $$ \frac{\sin 2x}{\cos 2x + \sin 270^{\circ}} = -\frac{\cos x}{\sin x} $$ - **Proof:** Simplify the left-hand side and show it equals $$ -\frac{\cos x}{\sin x} $$. - **Undefined When:** $$ \cos 2x + \sin 270^{\circ} = 0 $$ 4. **d)** $$ \frac{\cos 2x}{\sin x + \cos x} = \cos x - \sin x $$ - **Proof:** Simplify the left-hand side and show it equals $$ \cos x - \sin x $$. - **Undefined When:** $$ \sin x + \cos x = 0 $$ 5. **e)** $$ \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x $$ - **Proof:** Simplify the left-hand side and show it equals $$ \tan x $$. - **Undefined When:** $$ 1 + \cos x + \cos 2x = 0 $$ 6. **f)** $$ (1 - \tan A) \frac{\cos A}{\cos 2A} = \frac{1}{\cos A + \sin A} $$ - **Proof:** Simplify the left-hand side and show it equals $$ \frac{1}{\cos A + \sin A} $$. - **Undefined When:** $$ \cos A + \sin A = 0 $$ 7. **g)** $$ \frac{\sin 2x}{1 + \cos 2x} \cdot \frac{\sin 2x}{1 - \cos 2x} = 1 $$ - **Proof:** Simplify the left-hand side and show it equals 1. - **Undefined When:** $$ (1 + \cos 2x)(1 - \cos 2x) = 0 $$ 8. **h)** $$ \frac{\sin 3\theta}{\sin \theta} - \frac{\cos 3\theta}{\cos \theta} = 2 $$ - **Proof:** Simplify the left-hand side and show it equals 2. - **Undefined When:** $$ \sin \theta = 0 $$ or $$ \cos \theta = 0 $$ **Note:** Each identity is proven by simplifying the left-hand side to match the right-hand side, and the undefined values are determined by setting the denominators equal to zero.

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