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Question
Evaluate the following :
`(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)`
Solution
We have to find: `(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)`
Since `sec 70^@/(cosec 20^@) + sin 59^@/cos 31^@` and `sec (90^@ - theta) = cosec theta
So
`sec 70^@/(cosec 20^@) + sin 59^@/cos 31^@ = (sec (90^@ - 20^@))/(cosec 20^@) + (sin (90^@ - 31^@))/(cos 31^@)`
`= (cosec 20^@)/(cosec 20^@) + (cos 31^@)/(cos 31^@)`
1 + 1
= 2
So value of `(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)` is 2
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