Question

Solve: $\tan \theta +3\cot \theta =5\sec \theta$

Answer 1

Consider the given equation.

$\tan \theta +3\cot \theta =5sec\theta$

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

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

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

$1+2{\cos }^2\theta =5\sin \theta$

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

$1+2-2{\sin }^2\theta =5\sin \theta$

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

$2{\sin }^2\theta +6\sin \theta -\sin \theta -3=0$

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

We cannot consider $\sin \theta +3=0$.

Therefore,

$2\sin \theta -1=0$

$\sin \theta =\frac{1}{2}$

$\sin \theta =\sin \frac{\pi }{6}$

$\theta =n\pi +{(-1)}^n\frac{\pi }{6},n\in I$

Hence, this is the answer.