If cot x −tan x=sec x,

Question:

If $\cot x-\tan x=\sec x$, then, $x$ is equal to

(a) $2 n \pi+\frac{3 \pi}{2}, n \in Z$

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

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

(d) none of these.

Solution:

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

Given equation:

$\cot x-\tan x=\sec x$

$\Rightarrow \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=\frac{1}{\cos x}$

$\Rightarrow \frac{\cos ^{2} x-\sin ^{2} x}{\sin x \cos x}=\frac{1}{\cos x}$

$\Rightarrow \cos ^{2} x-\sin ^{2} x=\sin x$

$\Rightarrow\left(1-\sin ^{2} x\right)-\sin ^{2} x=\sin x$

$\Rightarrow 1-2 \sin ^{2} x=\sin x$

$\Rightarrow 2 \sin ^{2} x+\sin x-1=0$

$\Rightarrow 2 \sin ^{2} x+2 \sin x-\sin x-1=0$

$\Rightarrow 2 \sin x(\sin x+1)-1(\sin x+1)=0$

$\Rightarrow(\sin x+1)(2 \sin x-1)=0$

$\Rightarrow \sin x+1=0$ or $2 \sin x-1=0$

$\Rightarrow \sin x=-1$ or $\sin x=\frac{1}{2}$

Now,

$\sin x=-1 \Rightarrow \sin x=\sin \frac{3 \pi}{2} \Rightarrow x=m \pi+(-1)^{m} \frac{3 \pi}{2}, m \in Z$

and,

$\sin x=\frac{1}{2} \Rightarrow \sin x=\sin \frac{\pi}{6} \Rightarrow x=n \pi+(-1)^{n} \frac{\pi}{6}, n \in Z$

$\therefore x=n \pi+(-1)^{\mathrm{n}} \frac{\pi}{6}, \mathrm{n} \in \mathrm{Z}$

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