How do you solve #csc(x)+cot(x)=1 #?

Question

8924 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-12T17:08:46

# x=npi+(-1)^n*(pi/2), n in ZZ#.

Given that, #cscx+cotx=1..............(star_1)#.

We know that, #csc^2x-cot^2x=1#.

#"Factoring, "(cscx+cotx)(cscx-cotx)=1#.

#:. (1)(cscx-cotx)=1.........[because, (star_1)#.

# :. cscx-cotx=1.............................(star_2)#.

#(star_1)+(star_2) rArr 2cscx=2, or, cscx=1#.

#:. sinx=1=sin(pi/2)#.

Since, #sintheta=sinalpha rArr theta=npi+(-1)^nalpha, n in ZZ#,

#:. sinx=sin(pi/2) rArr x=npi+(-1)^n*(pi/2), n in ZZ#.