Question

If $0^0\le \theta \le {90}^0$, find the value of $\theta$ satisfying $3\tan \theta +\cot \theta =5\text{cosec}\theta$

• A. ${60}^0$
• B. ${90}^0$
• C. ${30}^0$
• D. None of these

Answer 1

The correct option is A ${60}^0$

$3\tan \theta +\cot \theta =5\text{cosec}\theta$
or $\frac{3\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{5}{\sin \theta }$
or $3{\sin }^2\theta +{\cos }^2\theta =5\cos \theta$
or $3(1-{\cos }^2\theta )+{\cos }^2\theta =5\cos \theta$
or $2{\cos }^2\theta +5\cos \theta -3=0$
or $(2\cos \theta -1)(\cos \theta +3)=0$
Therefore either
$2\cos \theta -1=0$ or $\cos \theta +3=0$
$\therefore \cos \theta =\frac{1}{2}$ or $-3$
i.e. possible value of $\theta ={60}^0$