$\begin{array}{1 1}(M+m)g\cot\theta\\(M+m)g\cos\theta\\(M+m)g\sin\theta\\(M+m)g\tan\theta\end{array} $

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Answer : $(M+m)g\tan\theta$

In this case M and m are moving with the same acceleration a. Note that a is along x-direction only

For M and m :

$F-N_2\sin \theta=Ma$--------(1)

$mg-N_2\cos\theta=0$$\Rightarrow N_2=\large\frac{mg}{\cos\theta}$------(2)

$N_2\sin \theta=ma$------(3)

From (2) and (3),

$a=g\tan\theta$

Substitute (3) into (1) obtains

$F = (M + m) a $

$\;\;\;= (M + m)g \tan\theta$

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