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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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Correct option is C

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,sec⁡(8θ)−1sec⁡(4θ)−1⋅tan⁡(2θ)tan⁡(8θ) is equal to

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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