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Question
Evaluate the following :
`(cot 40^@)/cos 35^@ - 1/2 [(cos 35^@)/(sin 55^@)]`
Solution
We have to find:
`(cot 40^@)/cos 35^@ - 1/2 [(cos 35^@)/(sin 55^@)]`
Since `cot(90^@ - theta) = tan theta` and `cos (90^@- theta) = sin theta`
`(cot 40^@)/(tan 50^@) - 1/2 ((cos 35^@)/(sin 55^@)) = cot(90^@ - 50^@)/ tan 50^@ - 1/2 (cos(90^@ - 55^@)/sin 55^@)`
`= tan 50^@/tan 50^@ - 1/2 ((sin 55^@)/(sin 55^@))`
`= 1 - 1/2`
`= 1/2`
So value of `(cot 40^@)/(tan 50^@) - 1/2 ((cos 35^@)/(sin 55^@)) " is " 1/2`
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