For 0<θ<π/4,sec⁡(8θ)−1sec⁡(4θ)−1⋅tan⁡(2θ)tan⁡(8θ) is equal to

sec⁡(8θ)−1sec⁡(4θ)−1⋅tan⁡(2θ)tan⁡(8θ)=1−cos⁡(8θ)1−cos⁡(4θ)⋅cos⁡(4θ)cos⁡(8θ)⋅sin⁡(2θ)cos⁡(2θ)⋅cos⁡(8θ)sin⁡(8θ)=2sin2⁡(4θ)2sin2⁡(2θ)cos⁡(4θ)sin⁡(2θ)cos⁡(2θ)sin⁡(8θ)=1

For 0<θ<π/4,sec⁡(8θ)−1sec⁡(4θ)−1⋅tan⁡(2θ)tan⁡(8θ) is equal to

sec⁡(8θ)−1sec⁡(4θ)−1⋅tan⁡(2θ)tan⁡(8θ)=1−cos⁡(8θ)1−cos⁡(4θ)⋅cos⁡(4θ)cos⁡(8θ)⋅sin⁡(2θ)cos⁡(2θ)⋅cos⁡(8θ)sin⁡(8θ)=2sin2⁡(4θ)2sin2⁡(2θ)cos⁡(4θ)sin⁡(2θ)cos⁡(2θ)sin⁡(8θ)=1

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For $0 < θ < π / 4, \frac{\sec (8 θ) - 1}{\sec (4 θ) - 1} \cdot \frac{\tan (2 θ)}{\tan (8 θ)}$ is equal to

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Detailed Solution

$\frac{\sec (8 θ) - 1}{\sec (4 θ) - 1} \cdot \frac{\tan (2 θ)}{\tan (8 θ)}$

$\begin{array}{l} = \frac{1 - \cos (8 θ)}{1 - \cos (4 θ)} \cdot \frac{\cos (4 θ)}{\cos (8 θ)} \cdot \frac{\sin (2 θ)}{\cos (2 θ)} \cdot \frac{\cos (8 θ)}{\sin (8 θ)} \\ = \frac{2 \sin^{2} (4 θ)}{2 \sin^{2} (2 θ)} \frac{\cos (4 θ) \sin (2 θ)}{\cos (2 θ) \sin (8 θ)} = 1 \end{array}$