Answer
$$\sigma_{x} = \sigma \cos^2 \theta$$
Work Step by Step
In the x-direction, the exerted force will be
$$F_{x}=F \cos \theta$$
Also, the area in the x-direction $A=A_x \cos \theta$
The mechanical stress equals the force divided by the area $A$ and in x-direction will be
$$\sigma_{x}=\frac{F_{x}}{A_{x}}$$
Now, substitute the terms of $A_x$ and $F_x$ to get the stress in the form
$$\sigma_{x}=\dfrac{F_{x}}{A_{x}} = \dfrac{F \cos \theta}{(A)/( \cos \theta)} = \dfrac{F}{A} \cos^2 \theta = \boxed{ \sigma \cos^2 \theta} $$
Where $\sigma = \dfrac{F}{A}$