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Question: $(n-1)$ 2) sec1. sec2 + sec2. sec3 + sec3. sec4 = sin...

(n−1)(n-1) 2) sec1. sec2 + sec2. sec3 + sec3. sec4 = sin

A

1^\circ

B

2^\circ

C

3^\circ

D

4^\circ

Answer

3^\circ

Explanation

Solution

The given equation is (n−1)(n-1) 2) sec⁡(1∘)sec⁡(2∘)+sec⁡(2∘)sec⁡(3∘)+sec⁡(3∘)sec⁡(4∘)=sin⁡(?)\sec(1^\circ) \sec(2^\circ) + \sec(2^\circ) \sec(3^\circ) + \sec(3^\circ) \sec(4^\circ) = \sin(?). The terms (n−1)(n-1) and "2)" are extraneous. We need to find the angle for the sine function.

Let the sum be S=sec⁡(1∘)sec⁡(2∘)+sec⁡(2∘)sec⁡(3∘)+sec⁡(3∘)sec⁡(4∘)S = \sec(1^\circ) \sec(2^\circ) + \sec(2^\circ) \sec(3^\circ) + \sec(3^\circ) \sec(4^\circ). We use the identity sec⁡Asec⁡B=tan⁡B−tan⁡Asin⁡(B−A)\sec A \sec B = \frac{\tan B - \tan A}{\sin(B-A)}. For each term, B−A=1∘B-A = 1^\circ. S=tan⁡(2∘)−tan⁡(1∘)sin⁡(1∘)+tan⁡(3∘)−tan⁡(2∘)sin⁡(1∘)+tan⁡(4∘)−tan⁡(3∘)sin⁡(1∘)S = \frac{\tan(2^\circ) - \tan(1^\circ)}{\sin(1^\circ)} + \frac{\tan(3^\circ) - \tan(2^\circ)}{\sin(1^\circ)} + \frac{\tan(4^\circ) - \tan(3^\circ)}{\sin(1^\circ)} This is a telescoping sum: S=1sin⁡(1∘)[(tan⁡(2∘)−tan⁡(1∘))+(tan⁡(3∘)−tan⁡(2∘))+(tan⁡(4∘)−tan⁡(3∘))]S = \frac{1}{\sin(1^\circ)} [(\tan(2^\circ) - \tan(1^\circ)) + (\tan(3^\circ) - \tan(2^\circ)) + (\tan(4^\circ) - \tan(3^\circ))] S=tan⁡(4∘)−tan⁡(1∘)sin⁡(1∘)S = \frac{\tan(4^\circ) - \tan(1^\circ)}{\sin(1^\circ)} Using tan⁡B−tan⁡A=sin⁡(B−A)cos⁡Acos⁡B\tan B - \tan A = \frac{\sin(B-A)}{\cos A \cos B}: S=sin⁡(4∘−1∘)cos⁡(1∘)cos⁡(4∘)sin⁡(1∘)=sin⁡(3∘)sin⁡(1∘)cos⁡(1∘)cos⁡(4∘)S = \frac{\sin(4^\circ - 1^\circ)}{\cos(1^\circ) \cos(4^\circ) \sin(1^\circ)} = \frac{\sin(3^\circ)}{\sin(1^\circ) \cos(1^\circ) \cos(4^\circ)} Using sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta, we have sin⁡(1∘)cos⁡(1∘)=12sin⁡(2∘)\sin(1^\circ)\cos(1^\circ) = \frac{1}{2}\sin(2^\circ). S=sin⁡(3∘)12sin⁡(2∘)cos⁡(4∘)=2sin⁡(3∘)sin⁡(2∘)cos⁡(4∘)S = \frac{\sin(3^\circ)}{\frac{1}{2}\sin(2^\circ) \cos(4^\circ)} = \frac{2\sin(3^\circ)}{\sin(2^\circ)\cos(4^\circ)}

The value of SS is approximately 3, which cannot be equal to sin⁡(?)\sin(?) as the maximum value of sine is 1. This indicates the question is ill-posed. However, if we interpret the question as asking for the most prominent angle in the simplified expression that could be related to the sine function, it is 3∘3^\circ appearing in the numerator of S=sin⁡(3∘)sin⁡(1∘)cos⁡(1∘)cos⁡(4∘)S = \frac{\sin(3^\circ)}{\sin(1^\circ) \cos(1^\circ) \cos(4^\circ)}. Assuming a flawed question where the intended answer is derived from the structure, 3∘3^\circ is the most plausible answer.