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Question : 9 of 160
Marks:
+1,
-0
Solution:
Consider the expression.
(| 1+cos‌−i‌sin‌ |
| 1+cos‌+i‌sin‌ |
)12 It is solved as,
(| 1+cos‌−i‌sin‌ |
| 1+cos‌+i‌sin‌ |
)12 =[| 2cos2−sin‌‌cos‌‌i |
| 2cos2+2‌sin‌‌cos‌‌i |
]12 =[| cos‌(cos‌−i‌sin‌) |
| cos‌(cos‌+i‌sin‌) |
]12 =[| (cos‌−i‌sin‌) |
| (cos‌+i‌sin‌) |
]12 =()12 Solve further,
(| 1+cos‌−i‌sin‌ |
| 1+cos‌+i‌sin‌ |
)12 =(e)12 =cos‌−i‌sin‌ =0−(−i) =i
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