A) \[\sin \beta \]
B) \[\cos \beta \]
C) \[\cot \beta \]
D) \[2\cos \beta \]
E) \[\cos ec\beta \]
Correct Answer: C
Solution :
Given,\[\alpha ,\beta ,\gamma \]are in AP \[\Rightarrow \] \[2\beta =\alpha +\gamma \] ...(i) \[\frac{\sin \alpha -\sin \gamma }{\cos \gamma -\cos \alpha }=\frac{2\cos \frac{\alpha +\gamma }{2}.\sin \frac{\alpha -\gamma }{2}}{2\sin \frac{\alpha -\gamma }{2}.\sin \frac{\alpha +\gamma }{2}}\] \[=\frac{2\cos \frac{2\beta }{2}}{2.\sin \frac{2\beta }{2}}\] [from Eq. (i)] \[=\frac{\cos \beta }{\sin \beta }=\cot \beta \]
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