Question

If $3\tan \theta +\cot \theta =5cosec\theta$, then $\theta =$ __ (in degrees) [$0^{\circ }<\theta <{90}^{\circ }$]

Answer 1

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

$\Rightarrow \sin \theta (3{\sin }^2\theta +{\cos }^2\theta )=5\sin \theta \cos \theta$

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

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

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

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

$\Rightarrow \cos \theta =-3$ or $\cos \theta =\frac{1}{2}$
Now, $\cos \theta =-3$ is not possible as $cos\theta \le 1$
Hence, $\cos \theta =\frac{1}{2}\Rightarrow \theta ={60}^{\circ }$