If cos A = −12/13 and cot B = 24/7, Find sin(A+B), cos(A+B) and tan(A+B)
Question
If \[ \cos A=-\frac{12}{13} \] and \[ \cot B=\frac{24}{7} \] where A lies in the second quadrant and B lies in the third quadrant, find:
(i) \(\sin(A+B)\)
(ii) \(\cos(A+B)\)
(iii) \(\tan(A+B)\)
Solution
Given:
\[ \cos A=-\frac{12}{13} \]
Using \[ \sin^2 A+\cos^2 A=1 \]
\[ \sin A=\sqrt{1-\left(-\frac{12}{13}\right)^2} \]
\[ =\sqrt{1-\frac{144}{169}} \]
\[ =\sqrt{\frac{25}{169}} \]
\[ \sin A=\frac{5}{13} \]
Since A lies in the second quadrant, sine is positive.
Therefore,
\[ \sin A=\frac{5}{13} \]
Also,
\[ \cot B=\frac{24}{7} \]
Therefore,
\[ \tan B=\frac{7}{24} \]
Using the Pythagorean triple:
\[ \sin B=\frac{7}{25}, \qquad \cos B=\frac{24}{25} \]
Since B lies in the third quadrant, both sine and cosine are negative.
Therefore,
\[ \sin B=-\frac{7}{25}, \qquad \cos B=-\frac{24}{25} \]
(i) Find \(\sin(A+B)\)
Using formula:
\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]
\[ =\frac{5}{13}\times\left(-\frac{24}{25}\right)+\left(-\frac{12}{13}\right)\times\left(-\frac{7}{25}\right) \]
\[ =-\frac{120}{325}+\frac{84}{325} \]
\[ =-\frac{36}{325} \]
Therefore,
\[ \boxed{\sin(A+B)=-\frac{36}{325}} \]
(ii) Find \(\cos(A+B)\)
Using formula:
\[ \cos(A+B)=\cos A\cos B-\sin A\sin B \]
\[ =\left(-\frac{12}{13}\right)\times\left(-\frac{24}{25}\right)-\frac{5}{13}\times\left(-\frac{7}{25}\right) \]
\[ =\frac{288}{325}+\frac{35}{325} \]
\[ =\frac{323}{325} \]
Therefore,
\[ \boxed{\cos(A+B)=\frac{323}{325}} \]
(iii) Find \(\tan(A+B)\)
Using formula:
\[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \]
First,
\[ \tan A=\frac{\sin A}{\cos A} \]
\[ =\frac{5/13}{-12/13} \]
\[ =-\frac{5}{12} \]
Also,
\[ \tan B=\frac{7}{24} \]
Now,
\[ \tan(A+B)=\frac{-\frac{5}{12}+\frac{7}{24}}{1-\left(-\frac{5}{12}\times\frac{7}{24}\right)} \]
\[ =\frac{-\frac{10}{24}+\frac{7}{24}}{1+\frac{35}{288}} \]
\[ =\frac{-\frac{3}{24}}{\frac{323}{288}} \]
\[ =-\frac{1}{8}\times\frac{288}{323} \]
\[ =-\frac{36}{323} \]
Therefore,
\[ \boxed{\tan(A+B)=-\frac{36}{323}} \]