Question

(cos47°÷sin43°)the whole square+(sin72°÷cos18°)the whole square-2cos square 45°

Answer 1

Answer:

Step-by-step explanation:

=(cos47/sin43)^2 + (sin72/cos18)^2 - 2cos^2 45                                                                          

=(sin90-47 / sin43)^2 + (cos90-72 / cos18) - 2cos^2 45

=(sin43/sin43)^2 + (cos18/cos18)^2 - 2(1/root2)^2

=1 + 1 - 2*1/2

= 2 - 1

= 1

Answer 2

Answer:

=(cos47/sin43)^2 + (sin72/cos18)^2 - 2cos^2 45                                                                          

=(sin90-47 / sin43)^2 + (cos90-72 / cos18) - 2cos^2 45

=(sin43/sin43)^2 + (cos18/cos18)^2 - 2(1/root2)^2

=1 + 1 - 2*1/2

= 2 - 1

= 1

Step-by-step explanation: