How do you evaluate #\sin 80^ { \circ } \sec 10^ { \circ }#?

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Answer 1 · 2018-06-07T20:13:45

#=1 #

Use #cos(90^circ - theta^circ ) = sin (theta^circ ) #

let #theta = 80^circ #

#=> cos(10^circ) = sin(80^circ) #

The question asks for:

#sin(80^circ) * 1/(cos(10^circ)) #

#=> sin(80^circ) * 1/ sin(80^circ) #

#= 1 #