# Answer to Question #22300 in Trigonometry for katie

Question #22300

prove that cotx+tanx=secx+cscx is equal

Expert's answer

cotx+tanx=secx+cscx

we know that

cot(x)=cos(x)/sin(x)

tan(x)=sin(x)/cos(x)

sec(x)=1/cos(x)

csc(x)=1/sin(x)

so we need to show thatcos(x)/sin(x)+sin(x)/cos(x)=1/cos(x)+1/sin(x)

we can multiple both parts sin(x)cos(x):

cos^2(x)+sin^2(x)=sin(x)+cos(x)

1=sin(x)+cos(x)

1=(sin(x)+cos(x))^2

1=1+2sin(x)cos(x)

sin(x)cos(x)=0

but for such x doesn't exist tan(x) or cot(x) so thisequality cant take place cotx+tanx=secx+cscx

we know that

cot(x)=cos(x)/sin(x)

tan(x)=sin(x)/cos(x)

sec(x)=1/cos(x)

csc(x)=1/sin(x)

so we need to show thatcos(x)/sin(x)+sin(x)/cos(x)=1/cos(x)+1/sin(x)

we can multiple both parts sin(x)cos(x):

cos^2(x)+sin^2(x)=sin(x)+cos(x)

1=sin(x)+cos(x)

1=(sin(x)+cos(x))^2

1=1+2sin(x)cos(x)

sin(x)cos(x)=0

but for such x doesn't exist tan(x) or cot(x) so thisequality cant take place cotx+tanx=secx+cscx

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