Question

solve 3tanθ + cotθ = 5cosecθ

Answer 1

3tanθ +cotθ = 5cosecθ
Squaring both sides,
$9tan^{2}$θ + $cot^{2}$θ + 6 = $25cosec^{2}$θ
$9tan^{2}$θ + $cot^{2}$θ + 6 = 25 + $25cot^{2}$θ
$9tan^{2}$θ - $24cot^{2}$θ - 19 = 0
Multiplying $tan^{2}$θ with all terms,
$9tan^{4}$θ - $19tan^{2}$θ - 24 = 0
Taking  $tan^{2}$θ as x, we get a quadratic equation
9x^2 - 19x -24 = 0
Solving this equation, we get
x = 3 or x = -0.88
As $tan^{2}$θ cannot be negative,
$tan^{2}$θ = 3
or tanθ = Square root of 3
or θ = 60 degrees or π/3

Answer 2

3tanθ +cotθ = 5cosecθ
Squaring both sides,
θ + θ + 6 = θ
θ + θ + 6 = 25 + θ
θ - θ - 19 = 0
Multiplying θ on both sides
θ - θ - 24 =
by Solving this equation, we get
x = 3 or x = -0.88
As θ cannot be negative,
θ = 3
or tanθ = Square root of 3
or θ = 60 degrees or π/3