Question

The sum of solutions of the equation \(\frac{\cos x}{1 + \sin x} = |\tan 2x|\), \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) - \left\{\frac{\pi}{4}, -\frac{\pi}{4}\right\}\) is :

• A. \(\frac{\pi}{10}\)
• B. \(-\frac{11\pi}{30}\)
• C. \(-\frac{7\pi}{30}\)
• D. \(-\frac{\pi}{15}\)

Hint:

Simplifying \(\frac{\cos x}{1+\sin x}\) to \(\tan(\pi/4 - x/2)\) is a common trick in JEE. Also, always verify that your solutions for \(x\) result in the correct sign for the terms inside an absolute value.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The equation involves trigonometric identities and absolute value. We first simplify the left-hand side using half-angle formulas and then solve for \(x\) by considering the cases for the absolute value function in the given interval.
Step 2: Key Formula or Approach:
1. \(\frac{\cos x}{1 + \sin x} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\).
2. If \(\tan A = \tan B\), then \(A = n\pi + B\).
Step 3: Detailed Explanation:
The left side is simplified as: \[ \frac{\cos x}{1 + \sin x} = \frac{\sin(\pi/2 - x)}{1 + \cos(\pi/2 - x)} = \frac{2\sin(\pi/4 - x/2)\cos(\pi/4 - x/2)}{2\cos^2(\pi/4 - x/2)} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \] So the equation is \(\tan(\pi/4 - x/2) = |\tan 2x|\). Since \(x \in (-\pi/2, \pi/2)\), \(\pi/4 - x/2 \in (0, \pi/2)\), so \(\tan(\pi/4 - x/2)>0\). This means the equation is always valid.
Case 1: \(\tan 2x>0\) \(\tan 2x = \tan(\pi/4 - x/2) \implies 2x = n\pi + \pi/4 - x/2 \implies \frac{5x}{2} = n\pi + \pi/4\). For \(n = 0\), \(x = \frac{\pi}{10}\). (Valid, as \(\tan 2(\pi/10)>0\)) For \(n = -1\), \(x = \frac{-3\pi}{4} \cdot \frac{2}{5} = -\frac{3\pi}{10}\). (Valid, as \(\tan 2(-3\pi/10) = \tan(-3\pi/5) = \tan(2\pi/5)>0\))
Case 2: \(\tan 2x<0\) \(-\tan 2x = \tan(\pi/4 - x/2) \implies \tan(-2x) = \tan(\pi/4 - x/2) \implies -2x = n\pi + \pi/4 - x/2 \implies -\frac{3x}{2} = n\pi + \pi/4\). For \(n = 0\), \(x = -\frac{\pi}{6}\). (Valid, as \(\tan 2(-\pi/6) = \tan(-\pi/3)<0\)) Sum of solutions: \[ S = \frac{\pi}{10} - \frac{3\pi}{10} - \frac{\pi}{6} = -\frac{2\pi}{10} - \frac{\pi}{6} = -\frac{\pi}{5} - \frac{\pi}{6} = -\frac{11\pi}{30} \]
Step 4: Final Answer:
The sum of the solutions is \(-\frac{11\pi}{30}\).