why here mgs not taken but mgs sin(alpha) ?

Dear student

Work done by any force is defined as product of force into displacement taken along force direction.
In above question, displacement of mass m is along the inclined plane.So, we have to consider the component of force which acts along the direction of the inclined plane.
$$\begin{gathered} In mg, mg\sin \alpha is the component which acts along the direction of displacement. \\ Whereas mg\cos \alpha is the component of force which acts perpendicular to the direction \\ of displacement. So contribution of mg\cos \alpha in work done is zero \\ whereas mg\sin \alpha fully contributes in the work done. \\ As displacement is s which is in direction of mg\sin \alpha \\ So, work done = mgs\times \sin \alpha \dots \dots work done = force\times displacement \end{gathered}$$

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