Question

Solve the following equations:
cot θ + cosec θ = `sqrt(3)`

Answer 1

cot θ + cosec θ = `sqrt(3)`

`cos theta/sin theta + 1/sin theta = sqrt(3)`, sin θ ≠ 0

`(cos theta + 1)/sin theta = sqrt(3)`, sin θ ≠ 0

1 + cos θ = `sqrt(3) sin theta`

`sqrt(3)sin theta - cos theta` = 1

Divide each term by 2

`sqrt(3)/2 sin theta - 1/2 cos theta = 1/2`

`sin  pi/3 * sin theta - cos  pi/3 * cos theta = 1/2`

`- (cos theta cos  pi/3 - sin theta sin  pi/3) = 1/2`

`cos (theta + pi/3) = - 1/2`

`cos (theta + pi/3) = cos (theta - pi/3)`

`cos (theta + pi/3) = cos ((3pi - pi)/3)`

`cos (theta + pi/3) = cos ((2pi)/3)`

The general solution is

`theta + pi/3 = 2"n"pi + (2pi)/3`, n ∈ Z

θ = `2"n"pi - pi/3 + (2pi)/3`, n ∈ Z

θ = `2"n"pi - pi/3 - (2pi)/3` or θ = `2"n"pi - pi/3 + (2pi)/3` 

θ = `2"n"pi - (3pi)/3` or θ = `2"n"pi + (2pi - pi)/3`

θ = `2"n"pi - pi` or θ = `2"n"pi + pi/3`, n ∈ Z

θ = `(2"n" - 1)pi` or θ = `2"n"pi + pi/3`, n ∈ Z

Since sin θ ≠ 0, θ = (2n – 1)π is not possible

∴ θ = `2"n"pi + pi/3`, n ∈ Z