Prove that (Cos2A Cos3A – Cos2A Cos7A + CosA Cos10A)/(Sin4A Sin3A – Sin2A Sin5A + Sin4A Sin7A) = Cot6A Cot5A

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Let’s simplify left hand side of this equation (Cos2A Cos3A – Cos2A Cos7A + CosA Cos10A)/(Sin4A Sin3A – Sin2A Sin5A + Sin4A Sin7A) = Cot6A Cot5A

$$\frac {Cos2A \text{ } Cos3A – Cos2A \text{ } Cos7A + CosA \text{ } Cos10A} {Sin4A \text{ } Sin3A – Sin2A \text{ } Sin5A + Sin4A \text{ } Sin7A}$$

Multiplying and dividing by 2

$$\frac {2\left(Cos2A \text{ } Cos3A – Cos2A \text{ } Cos7A + CosA \text{ } Cos10A\right)} {2\left(Sin4A \text{ } Sin3A – Sin2A \text{ } Sin5A + Sin4A \text{ } Sin7A\right)} \\ \text{ } \\ \frac {2 \text{ } Cos2A \text{ } Cos3A – 2 \text{ } Cos2A \text{ } Cos7A + 2 \text{ } CosA \text{ } Cos10A} {2 \text{ } Sin4A \text{ } Sin3A – 2 \text{ } Sin2A \text{ } Sin5A + 2 \text{ } Sin4A \text{ } Sin7A} \\$$

Simplifying further using formulas
2 CosA CosB = Cos(A + B) + Cos(A – B)

2 SinA SinB = Cos(A – B) – Cos(A + B)

$$\frac {2 \text{ } Cos2A \text{ } Cos3A – 2 \text{ } Cos2A \text{ } Cos7A + 2 \text{ } CosA \text{ } Cos10A} {2 \text{ } Sin4A \text{ } Sin3A – 2 \text{ } Sin2A \text{ } Sin5A + 2 \text{ } Sin4A \text{ } Sin7A} \\ \text{ } \\ \frac {Cos(2A + 3A) + Cos(2A – 3A) – [Cos(2A + 7A) + Cos(2A – 7A)] + Cos(A + 10A) + Cos(A – 10A)} {Cos(4A – 3A) – Cos(4A + 3A) – [Cos(2A – 5A) – Cos(2A + 5A)] + Cos(4A – 7A) – Cos(4A + 7A)} \\ \text{ } \\ \frac {Cos5A + Cos(- A) – [Cos9A + Cos(- 5A)] + Cos11A + Cos(- 9A)} {CosA – Cos7A – [Cos(- 3A) – Cos7A] + Cos(- 3A) – Cos11A} \\ \text{ } \\ \frac {Cos5A + Cos(- A) – Cos9A – Cos(- 5A) + Cos11A + Cos(- 9A)} {CosA – Cos7A – Cos(- 3A) + Cos7A + Cos(- 3A) – Cos11A} \\$$

For any value of θ, Cos(- θ) = Cosθ
Therefore

$$\frac {Cos5A + CosA – Cos9A – Cos5A + Cos11A + Cos9A} {CosA – Cos7A – Cos3A + Cos7A + Cos3A – Cos11A} \\ \text{ } \\ \frac {CosA + Cos11A} {CosA – Cos11A} \\$$

Using formulas
CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2

CosC – CosD = – 2 Sin(C + D)/2 Sin(C – D)/2

$$\frac {CosA + Cos11A} {CosA – Cos11A} \\ \text{ } \\ \frac{ 2 \text{ } Cos\left(\frac{A + 11A}{2}\right) \text{ } Cos\left(\frac{A – 11A}{2}\right)} {- 2 \text{ } Sin\left(\frac{A + 11A}{2}\right) \text{ } Sin\left(\frac{A – 11A}{2}\right)} \\ \text{ } \\ \frac{ 2 \text{ } Cos\left(\frac{12A}{2}\right) \text{ } Cos\left(\frac{- 10A}{2}\right)} {- 2 \text{ } Sin\left(\frac{12A}{2}\right) \text{ } Sin\left(\frac{- 10A}{2}\right)} \\ \text{ } \\ \frac {2 \text{ } Cos6A \text{ } Cos(- 5A)} {- 2 \text{ } Sin6A \text{ } Sin(- 5A)} \\ \text{ } \\ \frac {- Cos6A \text{ } Cos(- 5A)} {Sin6A \text{ } Sin(- 5A)} \\ \text{ } \\ – Tan6A \text{ } Cot(- 5A)$$

For any value of θ, Cot(- θ) = – Cotθ

$$Tan6A \text{ } Cot5A$$

Therefore (Cos2A Cos3A – Cos2A Cos7A + CosA Cos10A)/(Sin4A Sin3A – Sin2A Sin5A + Sin4A Sin7A) is equal to Tan6A Cot5A