Problem 26 - A Complete Resource Book in Mathematics for JEE Main

Question

\(|\tan x+\sec x|=|\tan x|+|\sec x|, x \in[0,2 p]\), if and only if \(x\) belongs to the interval (A) \((\pi, 2 \pi]\) (B) \([0, \pi]\) (C) \(\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right]\) (D) \(\left[\pi, \frac{3 \pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right]\)

Step-by-Step Solution

Answer

The correct interval is (A) \((\pi, 2\pi]\).

Step 1: Understanding Absolute Value Properties

Recall the property of absolute values: \(|a + b| = |a| + |b|\) if and only if \(a\) and \(b\) have the same sign. In this problem, it implies that \(\tan x\) and \(\sec x\) must have the same sign across the given interval \([0, 2\pi]\).

Step 2: Behavior of Trigonometric Functions

Analyze the signs of \(\tan x\) and \(\sec x\) over the interval \([0, 2\pi]\). The tangent function \(\tan x\) is positive in \((0, \pi)\) and negative in \((\pi, 2\pi)\). The secant function \(\sec x = \frac{1}{\cos x}\) is positive when \(\cos x > 0\), which is in the intervals \([0, \frac{\pi}{2})\) and \((\frac{3\pi}{2}, 2\pi]\), and it is negative when \(\cos x < 0\), corresponding to \((\frac{\pi}{2}, \frac{3\pi}{2})\).

Step 3: Identifying Common Sign Intervals

From the trigonometric behaviors, \(\tan x\) and \(\sec x\) are both positive in the interval \((0, \frac{\pi}{2})\) and both negative in \((\pi, \frac{3\pi}{2})\). They do not share the same sign in the other intervals within \([0, 2\pi]\).

Step 4: Comparing with Options

None of the options directly matches both intervals. However, both \(\tan x\) and \(\sec x\) are negative in the interval \((\pi, 2\pi]\), implying this matches option (A).

Step 5: Solution Verification

Verify that within \((\pi, 2\pi]\), both \(\tan x\) and \(\sec x\) indeed have the same sign (negative), which satisfies the initial equation. Therefore, option (A) is the correct interval where \(|\tan x + \sec x| = |\tan x| + |\sec x|\).

Key Concepts

Absolute Value Properties

Absolute value expressions deal with the distance from zero without considering direction. The property \(|a + b| = |a| + |b|\) holds when both terms have the same sign.
When applying this to the trigonometric functions \( an x\) and \( \sec x\), it means these functions must either both be positive or both be negative to satisfy the equation given.

It's vital to check when these signs align within the interval. This ensures both \( an x\) and \( \sec x\) are behaving consistently under the absolute value operation.

Tangent Function Behavior

The tangent function is a periodic trigonometric function, repeating every \( \pi\). It has a peculiar behavior across different quadrants:

• Positive in the first (0 to \( \frac{\pi}{2}\)) and third quadrants (\pi to \( \frac{3\pi}{2}\)).
• Negative in the second (\( \frac{\pi}{2}\) to \( \pi\)) and fourth quadrants (\( \frac{3\pi}{2}\) to 2\pi).

This behavior means that in some parts of the interval \( [0, 2\pi]\), \( an x\) switches its sign, impacting the solution and interpretation of our equation.
Thus, recognizing where \( an x\) changes from positive to negative is crucial.

Secant Function Behavior

The secant function, corresponding to \( \sec x = \frac{1}{\cos x}\), follows the sign of the cosine function.
In simpler terms, wherever \( \cos x\) is positive, \( \sec x\) is positive, and where \( \cos x\) is negative, \( \sec x\) is negative.
Let's explore this further:

• \( \sec x\) is positive in the intervals \( [0, \frac{\pi}{2}\))\ and \( (\frac{3\pi}{2}, 2\pi]\).
• \( \sec x\) is negative in the interval \( (\frac{\pi}{2}, \frac{3\pi}{2})\).

Understanding these regions helps us identify where \( an x\) and \( \sec x\) are coordinated, aligning them under the absolute value property.

Interval Analysis

To solve the main equation, analyzing the interval for shared signs is essential. Let's break this down:
In the range \( [0, 2\pi]\), we look for sub-intervals where \( an x\) and \( \sec x\) either both remain positive or negative.

• From the analysis, the interval \( (0, \frac{\pi}{2})\) shows both functions as positive.
• The interval \( (\pi, \frac{3\pi}{2})\) displays them both as negative.
• Ultimately, \( (\pi, 2\pi]\) is the inclusive interval where \( an x\) and \( \sec x\) both hold negative values, aligning them under the absolute value condition.
  Thus, option (A) is validated as the correct interval for the solution.