# Answer to Question #6648 in Trigonometry for Kyle Trace

Question #6648

cos betta+ sin betta tan betta= sec betta

Expert's answer

Let's denote betta as

x

cos(x)+sin(x)*tan(x)=cos(x)+sin(x)*(sin(x)/cos(x))=cos(x)+sin^2(x)/cos(x)=[cos(x)*cos(x)+sin^2(x)]/cos(x)=1/cos(x)

we use that tan(x)=sin(x)/cos(x) and cos^2(x)+sin^2(x)=1 for every x

x

cos(x)+sin(x)*tan(x)=cos(x)+sin(x)*(sin(x)/cos(x))=cos(x)+sin^2(x)/cos(x)=[cos(x)*cos(x)+sin^2(x)]/cos(x)=1/cos(x)

we use that tan(x)=sin(x)/cos(x) and cos^2(x)+sin^2(x)=1 for every x

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