Question

$3tan\theta$ + $cot\theta$ = 5 $cosec\theta$, $0\le \theta$ ≤ ${90}^{\circ }$. Then the value of $\theta$ is __${}^{\circ }$.

Answer 1

$\frac{3sin\theta }{cos\theta }$ + $\frac{cos\theta }{sin\theta }$ = $\frac{5}{sin\theta }$

$sin\theta$ ( $3si{n}^2\theta$ + $co{s}^2\theta$) = 5 $sin\theta cos\theta$

3(1 - $co{s}^2\theta$) + $co{s}^2\theta$ = 5 $cos\theta$

2$co{s}^2\theta$ + 5$cos\theta$ - 3 = 0

2$cos\theta$ [$cos\theta$ + 3] - ($cos\theta$ + 3) = 0

($cos\theta$ + 3) (2$cos\theta$ - 1) = 0

$cos\theta$ = -3 or $cos\theta$ = $\frac{1}{2}$

not possible as $cos\theta$ ≤ 1 $\theta$ = ${60}^{\circ }$