Question

If $\cot \mathrm{\theta }-\tan \mathrm{\theta }=\sec \mathrm{\theta }$, then, θ is equal to
(a) $2n\pi +\frac{3\pi }{2},n\in Z$

(b) $n\pi +{\left(-1\right)}^n\frac{\pi }{6},n\pi Z$

(c) $n\pi +\frac{\pi }{2},n\in Z$

(d) none of these.

Answer 1

(b) $n\pi +{\left(-1\right)}^n\frac{\pi }{6},n\in Z$
Given equation:

$$\begin{gathered} cot\theta -\tan \theta =sec\theta \\ \Rightarrow \frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos \theta }=\frac{1}{\cos \theta } \\ \Rightarrow \frac{{\cos }^2\theta -{\sin }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\cos \theta } \\ \Rightarrow {\cos }^2\theta -{\sin }^2\theta =\sin \theta \\ \Rightarrow (1-{\sin }^2\theta )-{\sin }^2\theta =\sin \theta \\ \Rightarrow 1-2{\sin }^2\theta =\sin \theta \\ \Rightarrow 2{\sin }^2\theta +\sin \theta -1=0 \\ \Rightarrow 2{\sin }^2\theta +2\sin \theta -\sin \theta -1=0 \\ \Rightarrow 2\sin \theta (\sin \theta +1)-1(\sin \theta +1)=0 \\ \Rightarrow (\sin \theta +1)(2\sin \theta -1)=0 \end{gathered}$$
$\Rightarrow \sin \theta +1=0$ or $2\sin \theta -1=0$
$\Rightarrow \sin \theta =-1$ or $\sin \theta =\frac{1}{2}$
Now,
$\sin \theta =-1\Rightarrow \sin \theta =\sin \frac{3\mathrm{\pi }}{2}\Rightarrow \theta =m\mathrm{\pi }+(-1{)}^m\frac{3\mathrm{\pi }}{2},m\in Z$
And,
$\sin \theta =\frac{1}{2}\Rightarrow \sin \theta =\sin \frac{\mathrm{\pi }}{6}\Rightarrow \theta =n\mathrm{\pi }+(-1{)}^n\frac{\mathrm{\pi }}{6},n\in Z$
∴ $\theta =n\mathrm{\pi }+(-1{)}^{\mathrm{n}}\frac{\mathrm{\pi }}{6},\mathrm{n}\in \mathrm{Z}$