Question

The values of $x$ in $[-2\pi ,2\pi ]$, for which the graph of the function $y=\sqrt{\frac{1+\sin x}{1-\sin x}}-secx$ and $y=-\sqrt{\frac{1+\sin x}{1-\sin x}}+secx$, coincide are

• A. $[-2\pi ,-\frac{3\pi }{2})\cup (\frac{3\pi }{2},2\pi ]$
• B. $(-\frac{3\pi }{2},\frac{\pi }{2})\cup (\frac{\pi }{2},\frac{3\pi }{2})$
• C. $(-\frac{\pi }{2},\frac{\pi }{2})$
• D. $[-2\pi ,2\pi ]-\{\pm \frac{\pi }{2},\pm \frac{3\pi }{2}\}$

Answer 1

Answer: (A), (C)

The correct options are
A $[-2\pi ,-\frac{3\pi }{2})\cup (\frac{3\pi }{2},2\pi ]$
C $(-\frac{\pi }{2},\frac{\pi }{2})$

Equating both the expression we get
$\sqrt{\frac{1+\sin x}{1-\sin x}}-\sec x=-\sqrt{\frac{1+\sin x}{1-\sin x}}+\sec x$
$2[\sqrt{\frac{1+\sin x}{1-\sin x}}-\sec x]=0$
$\sqrt{\frac{1+\sin x}{1-\sin x}}-\sec x=0$
$|\frac{1+\sin x}{\cos x}|-\sec x=0$
$|\sec x+\tan x|-\sec x=0$
$\sec x+\tan x-\sec x=0$
This implies
$\tan x=0$ or
$x=n\pi$
However only $n=2k$ satisfies the above equation.
Hence
$x=2k\pi$ where $k\in N$.
And
$-\sec x-\tan x-\sec x=0$
Or
$2\sec x+\tan x=0$
Or
$\frac{1}{\cos x}[2+\sin x]=0$
No solution ...
For the graphs coincide, $\sin \left(x\right)$ has to be a one one function in that given domain.
Hence the required domain is
$(\frac{-\pi }{2},\frac{\pi }{2})$
Or
$[-2\pi ,\frac{-3\pi }{2})\cup (\frac{3\pi }{2},2\pi ]$.