Question

$\left| {\tan x + \sec x} \right| = \left| {\tan x} \right| - \left| {\sec x} \right|$, $x \in \left[ {0,2\pi } \right]$ if and only if $x$ belongs to the interval
A) $\left[ {0,\pi } \right]$
B) $[0,\dfrac{\pi }{2}) \cup (\dfrac{\pi }{2},\pi ]$
C) $[\pi ,\dfrac{{3\pi }}{2}) \cup (\dfrac{{3\pi }}{2},2\pi ]$
D) None of these

Answer 1

Here, we are required to find the interval where $x$belongs such that $x \in \left[ {0,2\pi } \right]$ holds true. It is given that $\left| {\tan x + \sec x} \right| = \left| {\tan x} \right| - \left| {\sec x} \right|$. By our observation, we will find a relation between the terms present in the LHS and the RHS and then by solving the equation formed due to that relation, we will reach the desired answer.

Formula Used:
${\sec ^2}x = 1 + {\tan ^2}x$

Complete step by step solution:
According to the question
$\left| {\tan x + \sec x} \right| = \left| {\tan x} \right| - \left| {\sec x} \right|$
And, it is given that $x \in \left[ {0,2\pi } \right]$.
Since, $\left| {\tan x + \sec x} \right| = \left| {\tan x} \right| - \left| {\sec x} \right|$
Therefore, $\tan x.\sec x \leqslant 0$
This is because when we will open the modulus then since, the ‘minus’ sign is outside it, hence, $\sec x$will remain negative.
Hence, when we will multiply a positive term $\tan x$ with a negative $\sec x$, we will get a negative answer.
Also, as we can see LHS will be positive, hence, due to ‘equals to’ sign, RHS will also be positive, and this is possible if and only if,
$\left| {\tan x} \right| \geqslant \left| {\sec x} \right|$
Hence, when we will subtract $\sec x$from a greater term $\tan x$, then we will get a positive answer and hence, LHS$=$RHS.
Now, since, $\left| {\tan x} \right| \geqslant \left| {\sec x} \right|$
Squaring both sides,
$\Rightarrow {\tan ^2}x \geqslant {\sec ^2}x$………………………….(1)
As we know, ${\sec ^2}x = 1 + {\tan ^2}x$
Hence, substituting this in (1), we get
$\Rightarrow {\tan ^2}x \geqslant 1 + {\tan ^2}x$……………..(2)
For example, let ${\tan ^2}x = 0$
Therefore, from equation (2),
$\Rightarrow 0 \geqslant 1$
This is not possible.
Therefore, none of the given intervals can be true.

Hence, option D (none of these) is the correct option.

Note:
In this question, the modulus sign is given which means that when we remove it, we will get a non-negative answer of the term present inside the modulus. It is really important to compare the given trigonometric terms carefully to answer this question and we should take care of the fact that which term will remain positive and which will become negative as it could completely reverse our answer.