prove that 3tanθ+cotθ=5cosecθ
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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
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
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