Question

Solve $\cot \theta -\tan \theta =\sec \theta$.

Answer 1

$\frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos \theta }=\frac{1}{\cos \theta }$
or $\frac{\cos 2\theta }{\sin \theta \cos \theta }=\frac{1}{\cos \theta }$
$\therefore 1-2{\sin }^2\theta =\sin \theta$ $(\cos \theta \ne 0)$
or $2{\sin }^2\theta +\sin \theta -1=0$
or $(2\sin \theta -1)(\sin \theta +1)=0$
$\therefore \sin \theta =\frac{1}{2}$
$\therefore \theta =n\pi +(-1{)}^n\frac{\pi }{6}$
But $\sin \theta =-1$ implies $\cos \theta =0$ but $\cos \theta \ne 0$
Hence no value will be admissible when $\sin \theta =-1$.