Question

$3\tan \theta$ + $\cot \theta$ = 5cosec$\theta$. Solve for $\theta$, $0\le \theta \le 90$.

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Answer 1

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

$\frac{3{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{5}{\sin \theta }$
$3{\sin }^2\theta +{\cos }^2\theta =5\cos \theta$

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

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

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

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

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

Note that $\cos \theta$ = -3 is not possible as $-1\le \cos \theta \le 1$
Thus, $\theta$ = ${60}^{\circ }$