Given that $\tan A=\frac{12}{5}$, $\cos B=-\frac{3}{5}$ and that $A$ and $B$ are in the same quadrant, find the value of $\cos (A-B)$. $ \textbf{(A) }-\frac{63}{65}\qquad\textbf{(B) }-\frac{64}{65}\qquad\textbf{(C) }\frac{63}{65}\qquad\textbf{(D) }\frac{64}{65}\qquad\textbf{(E) }\frac{65}{63} $
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Tags: trigonometry
YESMAths
04.06.2014 14:32
SMOJ wrote:
Given that $\tan A=\frac{12}{5}$, $\cos B=-\frac{3}{5}$ and that $A$ and $B$ are in the same quadrant, find the value of $\cos (A-B)$.
$ \textbf{(A) }-\frac{63}{65}\qquad\textbf{(B) }-\frac{64}{65}\qquad\textbf{(C) }\frac{63}{65}\qquad\textbf{(D) }\frac{64}{65}\qquad\textbf{(E) }\frac{65}{63} $
Observe that $A$ and $B$ lie in the third quadrant.
So since $\cos{(A-B)}=\sin A\sin B+\cos A\cos B$, we have that $\cos (A-B)=\frac{63}{65}$
So the answer is $\textbf{(C) }\frac{63}{65}$
PlatinumFalcon
14.06.2014 23:40
Clarification: The "observation" noted uses the fact that (i) $\cos{B}$ is negative, (ii) $A$ and $B$ lie in the same quadrant, and (iii) $\tan{A}$ is positive. Since $\cos{B}$ is negative, and $A, B$ lie in the same quadrant, $\cos{A}$ must also be negative. Now $\tan{A}=\frac{\sin{A}}{\cos{A}}\in\mathbb{Z}^+$, so using the fact that $\cos{A}\in\mathbb{Z}^-$, we find that $\sin{A}\in\mathbb{Z}^-$. Hence, the only possible quadrant that we could be in is Quadrant III.