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Question
Prove that:
`cos 80^circ/(sin 10^circ) + cos 59^circ "cosec" 31^circ = 2`
Solution
`"LHS" = cos 80^circ/(sin 10^circ) + cos 59^circ "cosec" 31^circ `
`= (cos 80^circ)/cos(90^circ-10^circ) + sin (90^circ - 59^circ) "cosec" 31^circ`
`= 1 + "sin"31^circ`xx1/sin 31
= 1 + 1
= 2
= RHS
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