Question

The values of x in $[ - 2\pi ,2\pi ]$, for which the graph of the function $y = \sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} - \sec x$ and $y = - \sqrt {\dfrac{{1 - \sin x}}{{1 + \sin x}}} + \sec x$ , coincide are
A) $\left[ { - 2\pi , - \dfrac{{3\pi }}{2}} \right) \cup \left( {\dfrac{{3\pi }}{2},2\pi } \right]$
B) $\left( { - \dfrac{{3\pi }}{2}, - \dfrac{\pi }{2}} \right) \cup \left( {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right)$
C) $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$
D) \left[ { - 2\pi ,2\pi } \right] - \left\\{ { \pm \dfrac{\pi }{2}, \pm \dfrac{{3\pi }}{2}} \right\\}

Answer 1

We need to find the coinciding point of two functions which implies all the defined points of x for which the given functions satisfy under the restriction of x. As here the functions are provided, we will find the points of coincidence by equating both the functions and simplifying using appropriate trigonometric functions.
For simplifications we may need some of the basic trigonometric equations which are:
${\sin ^2}x + {\cos ^2}x = 1$
$\sec x = \dfrac{1}{{\cos x}}$
$\tan x = \dfrac{{\sin x}}{{\cos x}}$

Complete step-by-step answer:
Step 1: Equating both y values and simplifying with respect to x,
$\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} - \sec x = - \sqrt {\dfrac{{1 - \sin x}}{{1 + \sin x}}} + \sec x$
Multiplying the first term in LHS by $1 + \sin x$ and multiplying the first term in RHS by $1 - \sin x$,
$\sqrt {\dfrac{{1 + \sin x(1 + \sin x)}}{{1 - \sin x(1 + \sin x)}}} - \sec x = - \sqrt {\dfrac{{1 - \sin x(1 - \sin x)}}{{1 + \sin x(1 - \sin x)}}} + \sec x$
$\sqrt {\dfrac{{{{(1 + \sin x)}^2}}}{{(1 - \sin x)(1 + \sin x)}}} - \sec x = - \sqrt {\dfrac{{{{(1 - \sin x)}^2}}}{{(1 + \sin x)(1 - \sin x)}}} + \sec x$

Since $\sqrt {{{(1 + \sin x)}^2}} = 1 + \sin x$ and using the identity $(a + b)(a - b) = {a^2} - {b^2}$ in the denominators on either side.
$\dfrac{{1 + \sin x}}{{\cos x}} - \sec x = - \dfrac{{1 - \sin x}}{{\cos x}} + \sec x$ Since $1 - {\sin ^2}x = {\cos ^2}x$
$\dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}} - \sec x = - \dfrac{1}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}} + \sec x$
$\sec x + \tan x - \sec x = - \sec x - \tan x + \sec x$
$\sec x$ gets cancelled on both sides. Thus,
$\tan x = \tan x$
Implies both the functions are the same which is $\tan x$.
As we need x values in $[ - 2\pi ,2\pi ]$ and $\tan x$ is not defined for $\dfrac{\pi }{2}$ and multiples of $\dfrac{\pi }{2}$. So those not defined points should be eliminated.
Multiples of $\dfrac{\pi }{2}$which falls in the interval $[ - 2\pi ,2\pi ]$ are $\pm \dfrac{\pi }{2}$ and $\pm \dfrac{{3\pi }}{2}$.
Thus the required value of x is \left[ { - 2\pi ,2\pi } \right] - \left\\{ { \pm \dfrac{\pi }{2}, \pm \dfrac{{3\pi }}{2}} \right\\} which is option D.
Final answer:

Option D is correct. Required x value is \left[ { - 2\pi ,2\pi } \right] - \left\\{ { \pm \dfrac{\pi }{2}, \pm \dfrac{{3\pi }}{2}} \right\\}.

Note: Another way of approaching this problem is considering the data as a graphical representation. Then the point of coincidence will be the point where the graphs meet. Thus we can try the mathematical tools for graphical representation of given graphs. According to the restriction provided for the coinciding point x, we have to look for all values of x where these two graphs meet. Other basic trigonometric functions which are used for similar questions are,
${\tan ^2}x + {\cot ^2}x = 1$ , ${\sec ^2}x + \cos e{c^2}x = 1$ , $\cos ecx = \dfrac{1}{{\sin x}}$ , $\cot x = \dfrac{1}{{\tan x}}$ , $\cot x = \dfrac{{\cos x}}{{\sin x}}$ .