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Evaluate the following: (cos37^0)/(sin53...

Evaluate the following: `(cos37^0)/(sin53^0)`

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We have,
`(cos37^@)/(sin53^@)=(sin53^@)/(cos(90^@−53^@))=`(sin53^@)/(sin53^@)=1 ` [since, `cos(90^@−θ)=sinθ`]
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