A) 1, but not\[-1\]
B) \[-1\], but not 1
C) \[1\]or\[-1\]
D) 0
E) \[i\]
Correct Answer: C
Solution :
\[\because \]\[\tan (\alpha +\beta +\gamma )\] \[=\frac{\tan \alpha +\tan \beta +\tan \gamma -\tan \alpha \tan \beta \tan \gamma }{1-\tan \alpha \tan \beta -\tan \beta \tan \gamma -\tan \alpha \tan \gamma }\] \[=\frac{\tan \alpha \tan \beta \tan \gamma -\tan \alpha \tan \beta \tan \gamma }{1-\tan \alpha \tan \beta -\tan \beta \tan \gamma -\tan \alpha \tan \gamma }\] \[(\because \tan \alpha +\tan \beta +\tan \gamma =\tan \alpha \tan \beta \tan \gamma )\] \[\Rightarrow \] \[\tan (\alpha +\beta +\gamma )=0=\tan 0{}^\circ \] \[\Rightarrow \] \[\alpha +\beta +\gamma =0{}^\circ \,or\,\pi \] Now, \[xyz\] \[=(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta )\] \[(\cos \gamma +i\sin \gamma )\] \[=\cos (\alpha +\beta +\gamma )+i\sin (\alpha +\beta +\gamma )\] \[=\cos (0{}^\circ )+isin0{}^\circ =1\] Or \[\cos \pi +i\sin \pi =-1\] \[\therefore \] \[xyz=1\text{ }or\text{ }-1\]
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