Question

2cos^2a+1, при tga=n/4

Answer 1

(48+П²)/(16+П²)

sin²a/cos²a=П²/16

(1-cos²a)/cos²a=П²/16

16-16cos²a=П²cos²a

16=(16+П²)cos²a

cos²a=16/(16+П²)

2cos^2a+1=32/(16+П²)+1=(48+П²)/(16+П²)