3tanθ + cotθ = 5cosecθ. Solve for θ, 0≤θ≤90. __

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$3 tan θ$ + $cot θ$ = 5cosec $θ$ . Solve for $θ$ , $0 \leq θ \leq 90$ .

__

Options:

Answer: Option A
:

$\frac{3 sin θ}{cos θ} + \frac{cos θ}{sin θ}$ = $\frac{5}{sin θ}$
$\frac{3 {sin}^{2} θ + {cos}^{2} θ}{sin θ cos θ} = \frac{5}{sin θ}$
$3 {sin}^{2} θ + {cos}^{2} θ = 5 cos θ$
3(1 - ${cos}^{2} θ$ ) + ${cos}^{2} θ$ = 5 $cos θ$
2 ${cos}^{2} θ$ + 5 $cos θ$ - 3 = 0
2 $cos θ$   [ $cos θ$ + 3] - 1( $cos θ$ + 3) = 0
( $cos θ$ + 3) (2 $c o s θ$ - 1) = 0
$cos θ$ = -3     or      $cos θ$ = $\frac{1}{2}$
Note that $cos θ$ = -3 is not possible as $- 1 \leq cos θ \leq 1$
Thus, $θ$ = $60^{\circ}$

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