$\begin{array}{1 1}mg\sin\theta\\mg\cos\theta\\mg\tan\theta\\mg\cot\theta\end{array} $

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Answer : $mg\tan\theta$

To push the block up the inclined plane at a constant velocity, there must be no net force on the block along the direction of the incline.

Therefore the component of F along the incline must cancel the component of the weight along the incline.

This cancellation implies that $F cos\theta = mg \sin\theta$ or $F = mg \tan \theta$.

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