if tan of theta is 5/4 what is cos of theta?

So we have the opposite side (5) and the adjacent side (4) and we need the hypoteneuse.

So √(5² + 4²) = √41

And the cosine is the adjacent/hypoteneuse =    4/√(41)   or   4*√(41)/(41)

 Going off the old Soh Cah Toa, we can say Tan(θ)=Opposite/Adjactent (A/B in pic below).

From that, it is safe to assume that Opposite=5, Adjacent=4 (or some multiple thereof, that part doesn't matter).

Cos(θ)=Adjacent/Hypotenuse. We already know Adjacent, but we need the hypotenuse.

Another formula we need to remember is $${\mathtt{C}} = {\sqrt{{{\mathtt{A}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{B}}}^{{\mathtt{2}}}}}$$. We can calculate that the Hypotenuse is $${\sqrt{{{\mathtt{5}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{4}}}^{{\mathtt{2}}}}} = {\mathtt{6.403\: \!124\: \!237\: \!432\: \!848\: \!7}}$$ , or for simplicity$${\sqrt{{\mathtt{41}}}}$$.

With Hypotenuse figured out, we can get Cos(θ).

Cos(θ)=Adjactent/Hypotenuse=$${\frac{{\mathtt{4}}}{{\sqrt{{\mathtt{41}}}}}} = {\mathtt{0.624\: \!695\: \!047\: \!554\: \!424\: \!3}}$$

That is a really good answer anonymous!    

CPhill has just gone one little step further by rationalising the denominator.