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geometry

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Simplify $$\dfrac{\sin (A-B)}{\sin A \sin B} + \dfrac{\sin (B-C)}{\sin B \sin C} + \dfrac{\sin (C-A)}{\sin C \sin A}$$

Jun 19, 2020

1+0 Answers

#1
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Simplify

$$\dfrac{\sin (A-B)}{\sin A \sin B} + \dfrac{\sin (B-C)}{\sin B \sin C} + \dfrac{\sin (C-A)}{\sin C \sin A}$$

$$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\sin(A-B)}{\sin A \sin B} + \dfrac{\sin(B-C)}{\sin B \sin C} + \dfrac{\sin(C-A)}{\sin C \sin A} } \\\\ &=& \dfrac{\sin(A)\cos(B)-\cos(A)\sin(B)}{\sin A \sin B} + \dfrac{\sin(B)\cos(C)-\cos(B)\sin(C)}{\sin B \sin C} + \dfrac{\sin(C)\cos(A)-\cos(C)\sin(A)}{\sin C \sin A} \\\\ &=& \dfrac{\sin(A)\cos(B)}{\sin A \sin B}-\dfrac{\cos(A)\sin(B)}{\sin A \sin B} + \dfrac{\sin(B)\cos(C)}{\sin B \sin C}-\dfrac{\cos(B)\sin(C)}{\sin B \sin C} + \dfrac{\sin(C)\cos(A)}{\sin C \sin A}-\dfrac{\cos(C)\sin(A)}{\sin C \sin A} \\\\ &=& \dfrac{\cos(B)}{\sin B}-\dfrac{\cos(A)}{\sin A} + \dfrac{\cos(C)}{\sin C}-\dfrac{\cos(B)}{\sin B} + \dfrac{\cos(A)}{\sin A}-\dfrac{\cos(C)}{\sin C} \\\\ &=& \dfrac{\cos(B)}{\sin B}-\dfrac{\cos(B)}{\sin B} + \dfrac{\cos(C)}{\sin C}-\dfrac{\cos(C)}{\sin C} + \dfrac{\cos(A)}{\sin A}-\dfrac{\cos(A)}{\sin A} \\\\ &=& \mathbf{0} \\ \hline \end{array}$$

Jun 19, 2020