Question

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

 

पर्याय

• \[2 n\pi + \frac{3\pi}{2}, n \in Z\]

   

• \[n\pi + \left( - 1 \right)^n \frac{\pi}{6}, n \in Z\]

• \[n\pi + \frac{\pi}{2}, n \in Z\]

   

• none of these.

Answer 1

\[n\pi + \frac{\pi}{2}, n \in Z\]
Given equation:
\[cot x - \tan x = sec x\]
\[ \Rightarrow \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{1}{\cos x}\]
\[ \Rightarrow \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{1}{\cos x}\]
\[ \Rightarrow \cos^2 x - \sin^2 x = \sin x\]
\[ \Rightarrow (1 - \sin^2 x) - \sin^2 x = \sin x\]
\[ \Rightarrow 1 - 2 \sin^2 x = \sin x\]
\[ \Rightarrow 2 \sin^2 x + \sin x - 1 = 0\]
\[ \Rightarrow 2 \sin^2 x + 2 \sin x - \sin x - 1 = 0\]
\[ \Rightarrow 2 \sin x ( \sin x + 1) - 1 (\sin x + 1) = 0\]
\[ \Rightarrow (\sin x + 1) (2 \sin x - 1) = 0\]

\[\Rightarrow \sin x + 1 = 0\] 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,  

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

∴ \[x = n\pi + ( - 1 )^n \frac{\pi}{6} , n \in Z\]