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Question
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:
tan (A + B)
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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,
\[ \tan\left( A + B \right) = \frac{\sin\left( A + B \right)}{\cos\left( A + B \right)} = \frac{\frac{- 36}{325}}{\frac{323}{325}} = - \frac{36}{323}\]
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