A) \[2\]
B) \[-2\]
C) \[2i\]
D) \[-2i\]
Correct Answer: A
Solution :
Given, \[z=i{{\log }_{e}}(2-\sqrt{3})\] \[\because \] \[\cos z=\frac{{{e}^{iz}}+{{e}^{-iz}}}{2}\] \[=\frac{{{e}^{{{i}^{2}}{{\log }_{e}}(2-\sqrt{3})}}+{{e}^{{{\log }_{e}}(2-\sqrt{3})}}}{2}\] \[=\frac{{{e}^{{{\log }_{e}}{{(2-\sqrt{3})}^{-1}}}}+{{e}^{{{\log }_{e}}(2-\sqrt{3})}}}{2}\] \[=\frac{{{(2-\sqrt{3})}^{-1}}+(2-\sqrt{3})}{2}\] \[=\frac{1}{2}\left[ \frac{1}{2-\sqrt{3}}+2-\sqrt{3} \right]\] \[=\frac{1}{2}[2+\sqrt{3}+2-\sqrt{3}]=2\]
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