Question

3 tan theta+cot theta =5 cosec theta

Answer 1

We know, sin²θ = 1 - cos²θ use it here,
⇒3(1 - cos²θ) + cos²θ - 5cosθ = 0
⇒3 - 3cos²θ + cos²θ - 5cosθ = 0
⇒2cos²θ + 5cosθ - 3 = 0
⇒2cos²θ + 6cosθ - cosθ - 3 = 0
⇒2cosθ(cosθ + 3) - (cosθ+ 3) = 0
⇒(2cosθ - 1)(cosθ + 3) = 0
⇒cosθ = 1/2 and -3 but cosθ∈[ -1, 1 ] so, cosθ ≠ -3

Hence, cosθ = 1/2 = cosπ/3
θ = 2nπ ± π/3 [ by using general solution of trigonometric ]

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Answer 2

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HEY USER...
HERE IS UR ANS. ......

3tanθ + cotθ = 5cosecθ

as you know,
tanθ = sinθ/cosθ , cotθ = cosθ/sinθ and cosecθ = 1/sinθ use it here,

⇒3tanθ + cotθ = 5cosecθ
⇒3sinθ/cosθ + cosθ/sinθ = 5/sinθ
⇒(3sin²θ + cos²θ)/sinθ.cosθ = 5/sinθ
⇒3sin²θ + cos²θ = 5cosθ

We know, sin²θ = 1 - cos²θ use it here,
⇒3(1 - cos²θ) + cos²θ - 5cosθ = 0
⇒3 - 3cos²θ + cos²θ - 5cosθ = 0
⇒2cos²θ + 5cosθ - 3 = 0
⇒2cos²θ + 6cosθ - cosθ - 3 = 0
⇒2cosθ(cosθ + 3) - (cosθ+ 3) = 0
⇒(2cosθ - 1)(cosθ + 3) = 0
⇒cosθ = 1/2 and -3 but cosθ∈[ -1, 1 ] so, cosθ ≠ -3

Hence, cosθ = 1/2 = cosπ/3
θ = 2nπ ± π/3 [ by using general solution of trigonometric ]



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MRK AS BRAINLIEST

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