Solution
$$\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2+1$$
$$(\frac{1-\tan\theta}{1-\frac{1}{tan\theta}})^2+1$$
$$(\frac{1-\tan\theta}{\frac{tan\theta - 1}{tan\theta}})^2+1$$
$$(-tan\theta)^2 + 1$$
$$tan^2\theta + 1 = sec^2\theta$$
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