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

1 Answer
Jun 7, 2018

#=1 #

Explanation:

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 #