A) \[cosh\text{ }\alpha \text{ }cos\text{ }\beta \]
B) \[cos\text{ }\alpha \text{ }cos\text{ }\beta \]
C) \[cos\text{ }\alpha \text{ }cosh\text{ }\beta \]
D) \[sin\text{ }\alpha \text{ }sinh\text{ }\beta \]
Correct Answer: A
Solution :
\[\cosh (\alpha +i\beta )=\cos [i(\alpha +i\beta )]\] \[[\because \cosh \theta =\cos (i\theta )]\] \[=\cos (i\alpha -\beta )\] \[=\cos (i\alpha )\cos \beta +\sin (i\alpha )\sin \beta \] \[=\cos \alpha \cos \beta +i\sin \alpha .\sin \beta \] \[[\because \sin (i\theta )=i\sin (h\theta )]\] Hence, real part of\[\cosh (\alpha +i\beta )\]is\[\cosh \alpha \cos \beta \]
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