# If cos A = − 12 13 and cot B = 24 7 , where A lies in the second quadrant and B in the third quadrant, find the values of the following: cos (A + B) - Mathematics

Short Note

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,
$\cos\left( A + B \right) = \cos A \cos B - \sin A \sin B$
$= \frac{- 12}{13} \times \frac{- 24}{25} - \frac{5}{13} \times \frac{- 7}{25}$
$= \frac{288}{325} + \frac{35}{325}$
$= \frac{323}{325}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 8.2 | Page 19