Question

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

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

(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:

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