InterviewSolution
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InterviewSolution
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-sin(50^circ - theta) %2B cos(40^circ %2B theta) %2B (cos(40^circ)^2 %2B cos(50^circ)^2)/(sin(40^circ)^2 %2B sin(50^circ)^2) |
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Answer» Cos(40+x)-Sin(50-x)+ Cos^2 40+ cos^2 50/ Sin^2 40+ Sin^2 50=Cos(40-90)-Sin(40-90) + cos^2(40+50)+ Cos^2(50+40)/Sin^2(40+50)+Sin(40+50)= Sin50-Sin50+ cos^2(90)+Sin^2(90)/ Sin^2 (90)+ Sin^2(90)=1-1+1/1=0 |
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