A) \[2\text{ }cos(\alpha +\beta )\]
B) \[2\text{ }\sin (\alpha +\beta )\]
C) \[2\text{ }cos(\alpha -\beta )\]
D) \[4\cos \alpha \cos \beta \]
Correct Answer: A
Solution :
Given, \[2\cos \alpha =a+\frac{1}{a}\] \[\Rightarrow \] \[{{a}^{2}}-2a\cos \alpha +1=0\] ...(i) and \[2\cos \beta =b+\frac{1}{b}\] \[\Rightarrow \] \[{{b}^{2}}-2b\cos \beta +1=0\] ...(ii) From Eq. (i), \[a=\frac{2\cos \alpha \pm \sqrt{4{{\cos }^{2}}\alpha -4}}{2}\] \[\Rightarrow \] \[a=\frac{2\cos \alpha \pm 2i\sin \alpha }{2}\] \[\Rightarrow \] \[a=\cos \alpha \pm i\sin \alpha \] Similarly, \[b=cos\text{ }\beta \pm i\text{ }sin\text{ }\beta \] \[\therefore \]\[ab=(\cos \alpha \pm i\sin \alpha )(\cos \beta \pm i\sin B)\] \[=\cos (\alpha +\beta )\pm i\sin (\alpha +\beta )\] and \[\frac{1}{ab}=\cos (\alpha +\beta )\mp i\sin (\alpha +\beta )\] Hence, \[ab+\frac{1}{ab}=2\cos (\alpha +\beta )\]
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