Question #04552

1 Answer
Y
May 12, 2017

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

Explanation:

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#.

Related questions
Impact of this question
800 views around the world
You can reuse this answer
Creative Commons License