If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:

cos (A + B)

#### Solution

Given:

\[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\]

A lies in thesecond quadrant and B lies in the third quadrant .

We know that sine function is positive in thesecond quadrant and in thethird quadrant, both sine and cosine functions are negative.

Therefore,

\[\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left( \frac{- 12}{13} \right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}\]

\[\sin B = - \frac{1}{\sqrt{1 + \cot^2 B}} = - \frac{1}{\sqrt{1 + \left( \frac{24}{7} \right)^2}} = \frac{- 1}{\sqrt{1 + \frac{576}{49}}} = \frac{- 1}{\sqrt{\frac{625}{49}}} = \frac{- 7}{25}\]

\[\cos B = - \sqrt{1 - \sin^2 B} = - \sqrt{1 - \left( \frac{- 7}{25} \right)^2} = - \sqrt{1 - \frac{49}{625}} = - \sqrt{\frac{576}{625}} = - \frac{24}{25}\]

Now,

\[\cos\left( A + B \right) = \cos A \cos B - \sin A \sin B\]

\[ = \frac{- 12}{13} \times \frac{- 24}{25} - \frac{5}{13} \times \frac{- 7}{25}\]

\[ = \frac{288}{325} + \frac{35}{325}\]

\[ = \frac{323}{325}\]