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**Hint:**The cross product of two vectors is also a vector quantity The cross product a$ \times $b is defined as a vector c that is perpendicular to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.

$A \times B = \left | A \right |\left | B \right |\operatorname{Sin} \theta n$

Where

$\left | A \right |$ is the length of vector A

$\left | B \right |$ is the length of vector B

Is the angle between A & B

N is the unit vector perpendicular to the plane containing A and B

**Complete step by step solution:**

Refer to the following image

Now, consider two unit vectors in the X-Y plane as follows :

$\hat a$→ unit vector inclined with the positive direction of X-axis at angles A

\[\hat b\]→ unit vector inclined with the positive direction of X-axis at angles 90-B, where $90 - B > A$

The angle between these two vectors becomes

$\theta = 90 - A - B = 90 - \left( {A + B} \right)$

Now writing in vector form we get

$\hat a = \operatorname{Cos} A\hat i + \operatorname{Sin} A\hat j$

$\hat b = \operatorname{Cos} \left( {90 - B} \right)\hat i + \operatorname{Sin} \left( {90 - B} \right)\hat j$

$\hat b = \operatorname{Sin} B\hat i + \operatorname{Cos} B\hat j$

Now

Taking the cross product of the above two vectors i.e.,

$\hat a \times \hat b = \left( {\operatorname{Cos} A\hat i + \operatorname{Sin} A\hat j} \right) \times \left( {\operatorname{Sin} B\hat i + \operatorname{Cos} B\hat j} \right)$

$\because A \times B = \left | A \right |\left | B \right |\operatorname{Sin} \theta \hat k$

$\therefore \left | {\hat a} \right | \left | {\hat b} \right |\operatorname{Sin} \theta \hat k = \operatorname{Cos} A\operatorname{Cos} B\left( {\hat i \times \hat j} \right) + \operatorname{Sin} A\operatorname{Sin} B(\hat j \times \hat i)$ Applying Properties of unit vectors $\hat i,\hat j,\hat k$

$\hat i \times \hat j = \hat k$ $\hat j \times \hat i = - \hat k$

$\hat i \times \hat i = null$ $\hat j \times \hat j = null$ and

$\left | A \right |$=1 and $\left | B \right |$=1 as both are unit vectors

Also substituting the value of the angle between the vectors, $\theta = 90 - \left( {A + B} \right)$

Finally, we get,

\[\operatorname{Sin} \left( {90 - \left( {A + B} \right)} \right)\hat k = \operatorname{Cos} A\operatorname{Cos} B\hat k + \operatorname{Sin} A\operatorname{Sin} B\hat k\]

$\therefore \operatorname{Cos} \left( {A + B} \right) = \operatorname{Cos} A\operatorname{Cos} B - \operatorname{Sin} A\operatorname{Sin} B$

**Note:**

The Cross product is a vector quantity

If two vectors are parallel to each other then their cross product will be zero since $\operatorname{Sin} 0^\circ = 0$

The dot product is a scalar quantity

If two vectors are perpendicular their dot product will be zero since $\operatorname{Cos} 90^\circ = 0$

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