SOLUTION: How do you prove that this is an identity? sinX/cosX + cosX/sinX = secX cscX I've been thinking that in the end both sides should somehow end up with 1/(cosX)(sinX) but I rea

Algebra ->  Trigonometry-basics -> SOLUTION: How do you prove that this is an identity? sinX/cosX + cosX/sinX = secX cscX I've been thinking that in the end both sides should somehow end up with 1/(cosX)(sinX) but I rea      Log On


   

Question 602403: How do you prove that this is an identity?
sinX/cosX + cosX/sinX = secX cscX
I've been thinking that in the end both sides should somehow end up with 1/(cosX)(sinX) but I really don't know how to do that since the left side has addition and not multiplication so maybe I'm wrong? IDK I'm so confused.
Thank you to whoever answers this!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
(sin(x))/(cos(x))+(cos(x))/(sin(x)) = sec(x)csc(x)

(sin(x)sin(x))/(sin(x)cos(x))+(cos(x))/(sin(x)) = sec(x)csc(x)

(sin^2(x))/(sin(x)cos(x))+(cos(x))/(sin(x)) = sec(x)csc(x)

(sin^2(x))/(sin(x)cos(x))+(cos(x)cos(x))/(sin(x)cos(x)) = sec(x)csc(x)

(sin^2(x))/(sin(x)cos(x))+(cos^2(x))/(sin(x)cos(x)) = sec(x)csc(x)

(sin^2(x)+cos^2(x))/(sin(x)cos(x)) = sec(x)csc(x)

(1)/(sin(x)cos(x)) = sec(x)csc(x)

(1/sin(x))*(1/cos(x))= sec(x)csc(x)

csc(x)*sec(x)= sec(x)csc(x)

sec(x)csc(x)= sec(x)csc(x)


This verifies the identity.