Answers

The Brainliest Answer!
2015-02-21T13:55:34+05:30
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





1 5 1
Please choose my answer as the best answer
2015-02-22T13:28:48+05:30
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
0