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If $\tan \theta = \frac{a}{b},$ then $\frac{\sin \theta}{\cos^8 \theta} + \frac{\cos \theta}{\sin^8 \theta} = $

Prove that:
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If $\theta$ lies in the second quadrant,then the value of $\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} + \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}$ is:

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