જો $\frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{a + b \tan^2 2\theta}{1 + c \tan^2 2\theta + d \tan^4 2\theta}$ (જ્યાં $\theta \neq \frac{n\pi}{16}, n \in I$),તો $(a - b + c - d)$ ની કિંમત શોધો.
- A$0$
- B$1$
- C$7$
- D$13$
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