Question

MPLE |20| Prove that \[ \begin{aligned} \frac{\cos 2 A \cos 3 A-\cos 2 A \cos 7 A+\cos A \cos 10 A}{\sin 4 A \sin 3 A-\sin 2 A \sin 5 A+\sin 4 A \sin 7 A} & \\ &=\cot 6 A \cdot \cot 5 A \end{aligned} \]

Answer 1

\(\frac{cos\,2A\,cos\,3A-cos\,2A\,cos\,7A+cos\,A\,cos\,10A}{sin\,4A\,sin\,3A-sin\,2A\,sin\,5A+sin\,4A\,sin\,7A}\)

\(=\frac{2\,cos\,2A\,cos\,3A-2\,cos\,2A\,cos\,7A+2\,cos\,A\,cos\,10A}{2\,sin\,4A\,sin\,3A-2\,sin\,2A\,sin\,5A+2\,sin\,4A\,sin\,7A}\)

\(=\frac{cos(2A+3A)+cos(2A-3A)-(cos(2A+7A)\\\;\;\;\;\;\;+cos(2A-7A))+cos(A+10A)+cos(A-10A)}{cos(4A-3A)-cos(4A+3A)-(cos(2A-5A)\\\;\;\;\;\;\;-cos(2A+5A))+cos(4A-7A)-cos(4A+7A)}\)

(∵ 2 cos A cos B = cos (A+B) + cos (A-B) & 2 sin A sin B = cos (A-B) - cos (A+B))

\(=\frac{cos\,5A+cos(-A)-cos\,9A-cos(-5A)+cos(11A)+cos(-9A)}{cos(-A)-cos\,7A-cos(-3A)+cos\,7A+cos(-3A)-cos\,11A}\)

\(=\frac{cos\,5A+cos\,A-cos\,9A-cos\,5A+cos\,11A+cos\,9A}{cos\,A-cos\,7A-cos\,3A+cos\,7A+cos\,3A-cos\,11A}\)

(∵ cos (-θ) = cos θ)

\(=\frac{cos\,A+cos\,11A}{cos\,A-cos\,11A}\)

\(=\cfrac{2\,cos\,\frac{12A}A\,cos\,(\frac{-10A}2)}{2\,sin\,\frac{12A}A\,sin\,\frac{10A}2}\)

(∵ cos C + cos D = 2 cos (C+D/2) cos (C-D/2) and cos C - cos D = 2 sin (C+D/2) sin (D-C/2))

\(=\frac{cos\,6A\,cos\,5A}{sin\,6A\,sin\,5A}\) (∵ cos (-θ) = cos θ)

= cot 6A cot 5A

Hence proved.