So I understand where the formula for cos between two vectors comes from. However, if i have a paralelogram of basis vec(a) , vec(b) , and i want to find out the angle between its diagonals starting only from the fact that norm(vec(a))=3 and norm(vec(b))=2 and the angle between them pi/3 , how would i go about it?

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$$\left[\begin{array}{cccc}1& 3& 0& -4\\ 2& 6& 0& -8\end{array}\right]$$

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b) ${\displaystyle f\left(n\right)=\sqrt{n^2+1}}$. 
c) ${\displaystyle f\left(n\right)=\frac{1}{n^2-4}}$.

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The solution set of $\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=3\cot <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$ is

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