1.

40^circ %2B theta - sin(50^circ - theta) %2B (cos(40^circ)^2 %2B cos(50)^2)/(sin(40^circ)^2 %2B sin(50)^2)

Answer»

cos(40+90)- sin(50-90)+ cos^40+ cos^2(90-50)/sin^2(90-40)+ sin^2(50)= cos^2 (40)+ sin^2(40)+ sin^2(40)+ sin^2(50)/ cos^2(50)+ cos^2(50)=1+1×1

1 is the correct answer of the given question.


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