Please.solve example 51

Please.solve example 51 sin 4A cos3A + sin4A cos A cos4A cos3A + cos4A cosA sin4A(cos3A+ osA) cos4A(cos3A cosA) RHS sin(0 + a) 1 -m then prove that tan — -0 4 cos(0 - a) •m First Method sin(0 + a) I-m cos(0 - (1) 1 + m .tan 4

\frac{\sin (θ + α)}{\cos (θ - α)} = \frac{1 - m}{1 + m} Using componendo and divindo, \frac{1 - m + 1 + m}{1 - m - 1 - m} = \frac{\sin (θ + α) + \cos (θ - α)}{\sin (θ + α) - \cos (θ - α)} \frac{2}{- 2 m} = \frac{\sin (θ + α) + \cos (θ - α)}{\sin (θ + α) - \cos (θ - α)} m = \frac{\cos (θ - α) - \sin (θ + α)}{\sin (θ + α) + \cos (θ - α)} = \frac{cosθcosα + sinθsinα - sinθcosα - cosθsinα}{sinθcosα + cosθsinα + cosθcosα + sinθsinα} = \frac{cosθ (cosα - sinα) - sinθ (cosα - sinα)}{sinθ (cosα + sinα) + cosθ (cosα + sinα)} = \frac{(cosα - sinα)}{(cosα + sinα)} \times \frac{cosθ - sinθ}{cosθ + sinθ} = \frac{1 - tanα}{1 + 1 \times tanα} \times \frac{1 - tanθ}{1 + 1 \times tanθ} = \frac{\tan \frac{π}{4} - tanα}{1 + \tan \frac{π}{4} tanα} \times \frac{\tan \frac{π}{4} - tanθ}{1 + \tan \frac{π}{4} tanθ} = \tan (\frac{π}{4} - α) \tan (\frac{π}{4} - θ)