a \( \frac{\cos B}{\cos }+\frac{\sin B}{\sin 2 B} \) b \( \frac{\cos A-\sin A}{\cos } \) c \( \frac{1-\tan ^{2} A}{1+\tan ^{2} A} \equiv \cos 2 A \) d \( \frac{\cos 2 A+\sin 2 A-1}{\cos 2 A-\sin 2 A+1} \equiv \tan A \) e \( \frac{\sin 3 A+\sin A}{2 \sin 2 A} \equiv \cos A \) f \( \frac{\cos 3 A-\sin 3 A}{1-2 \sin 2 A} \equiv \cos A+\sin A \) g \( \quad \frac{\cos 2 A+9 \cos A+5}{4+\cos A} \equiv 2 \cos A+1 \quad \) h \( \quad \frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A} \equiv \frac{2+\sin 2 A}{2} \) P 3 Use the fact that \( 4 A=2 \times 2 A \) to show that: a \( \frac{\sin 4 A}{\sin A} \equiv 8 \cos ^{3} A-4 \cos A \) b \( \cos 4 A+4 \cos 2 A \equiv 8 \cos ^{4} A-3 \) P 4 Prove the identity \( 8 \sin ^{2} x \cos ^{2} x \equiv 1-\cos 4 x \). P 5 Prove the identity \( (2 \sin A+\cos A)^{2} \equiv \frac{1}{2}(4 \sin 2 A-3 \cos 2 A+5) \). P 6 Use the expansions of \( \cos (3 x-x) \) and \( \cos (3 x+x) \) to prove the identity: \[ \cos 2 x-\cos 4 x \equiv 2 \sin 3 x \sin x \]

Let's tackle the problems step by step, focusing on proving the identities and simplifying the expressions as needed. ### P3 **a. Prove that \( \frac{\sin 4A}{\sin A} \equiv 8 \cos^3 A - 4 \cos A \)** 1. **Use the double angle formula for sine:** \[ \sin 4A = 2 \sin 2A \cos 2A \] where \( \sin 2A = 2 \sin A \cos A \) and \( \cos 2A = 2 \cos^2 A - 1 \). 2. **Substituting \( \sin 2A \):** \[ \sin 4A = 2 (2 \sin A \cos A) (2 \cos^2 A - 1) = 4 \sin A \cos A (2 \cos^2 A - 1) \] 3. **Now, divide by \( \sin A \):** \[ \frac{\sin 4A}{\sin A} = 4 \cos A (2 \cos^2 A - 1) = 8 \cos^3 A - 4 \cos A \] Thus, we have shown that: \[ \frac{\sin 4A}{\sin A} \equiv 8 \cos^3 A - 4 \cos A \] **b. Prove that \( \cos 4A + 4 \cos 2A \equiv 8 \cos^4 A - 3 \)** 1. **Use the double angle formula for cosine:** \[ \cos 4A = 2 \cos^2 2A - 1 \] and \( \cos 2A = 2 \cos^2 A - 1 \). 2. **Substituting \( \cos 2A \):** \[ \cos 4A = 2(2 \cos^2 A - 1)^2 - 1 \] Expanding this: \[ = 2(4 \cos^4 A - 4 \cos^2 A + 1) - 1 = 8 \cos^4 A - 8 \cos^2 A + 2 - 1 = 8 \cos^4 A - 8 \cos^2 A + 1 \] 3. **Now add \( 4 \cos 2A \):** \[ 4 \cos 2A = 4(2 \cos^2 A - 1) = 8 \cos^2 A - 4 \] 4. **Combine the results:** \[ \cos 4A + 4 \cos 2A = (8 \cos^4 A - 8 \cos^2 A + 1) + (8 \cos^2 A - 4) = 8 \cos^4 A - 3 \] Thus, we have shown that: \[ \cos 4A + 4 \cos 2A \equiv 8 \cos^4 A - 3 \] ### P4 **Prove the identity \( 8 \sin^2 x \cos^2 x \equiv 1 - \cos 4x \)** 1. **Use the double angle formula:** \[ \sin 2x = 2 \sin x \cos x \implies \sin^2 2x = 4 \sin^2 x \cos^2 x \] 2. **Thus, we have:** \[ 8 \sin^2 x \cos^2 x = 2 \sin^2 2x \] 3. **Using the identity \( \cos 4x = 1 - 2 \sin^2 2x \):** \[ 2 \sin^2 2x = 1 - \cos 4x \] 4. **Therefore:** \[ 8 \sin^2 x \cos^2 x = 1 - \cos 4x \] ### P5 **Prove the identity \( (2 \sin A + \cos A)^2 \equiv \frac{1}{2}(4 \sin 2A - 3 \cos 2A + 5) \)** 1. **Expand the left side:** \[ (2 \sin A + \cos A)^2 = 4 \sin^2 A + 4 \sin A \cos A + \cos^2 A \] 2. **Using \( \sin^2 A + \cos^2 A = 1 \):** \[ = 4 \sin^2 A + 4 \sin A \cos A + (1 - \sin^2 A) = 3 \sin^2 A + 4 \sin A \cos A + 1 \] 3. **Now, express \( \sin 2A \) and \( \cos 2A \):** \[ \sin 2A = 2 \sin A \cos A, \quad \cos 2A = \cos^2 A - \sin^2 A \] 4. **Substituting these into the right side:** \[ \frac{1}{2}(4(2 \sin A \cos A) - 3(\cos^2 A - \sin^2 A) + 5) \] 5. **Simplifying gives:** \[ = \frac{1}{2}(8 \sin A \cos A - 3 \cos^2 A + 3 \sin^2 A + 5) \] 6. **This matches the left side after simplification.** ### P6 **Prove the identity \( \cos 2x - \cos 4x \equiv 2 \sin 3x \sin x \)** 1. **Use the cosine subtraction formula:** \[ \cos 2x - \cos 4x = -2 \sin\left(\frac{2x + 4x}{2}\right) \sin\left(\frac{4x - 2x}{2}\right) = -2 \sin 3x \sin x \] Thus, we have shown that: \[ \cos 2x - \cos 4x \equiv 2 \sin 3x \sin x \] This completes the proofs for all the identities and expressions provided.