Question #04552

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Answer 1 · 2017-05-12T04:20:55

Use #sin(theta)=cos(90^@-theta)#. Answer: #2#

Evaluate #(sin39^@/cos51^@)^2+(cos51^@/sin39^@)^2#

Here, it is important to note that #sin(theta)=cos(90^@-theta)# and #cos(theta)=sin(90^@-theta)#

Therefore, by substitution:

#(sin39^@/cos51^@)^2+(cos51^@/sin39^@)^2#

#=(cos(90^@-39^@)/cos51^@)^2+(cos51^@/cos(90^@-39^@))^2#

#=(cos51^@/cos51^@)^2+(cos51^@/cos51^@)^2#

#=1^2+1^2#

#=2#

Therefore, our answer is #2#.