Question

Number of solutions of equation \((\sin \theta+\cos \theta) \tan \theta=\sqrt{\sec ^{2} \theta-1}\) in \([0,2 \pi]\) are (1) 1 (2) 2 (3) 3 (4) 4

Short answer

4

Step-by-step solution

Simplify the Right-Hand Side

The right-hand side of the equation involves the expression \[ \sqrt{\sec^2 \theta - 1} \]. Recall that \[ \sec^2 \theta = 1 + \tan^2 \theta \], so \[ \sec^2 \theta - 1 = \tan^2 \theta \]. Thus, the right-hand side simplifies to \[ \sqrt{\tan^2 \theta} = |\tan \theta| \].

Address Absolute Value

Now consider the equation \[ (\sin \theta + \cos \theta) \tan \theta = |\tan \theta| \]. There are two cases to consider: \[ \tan \theta = |\tan \theta| \] and \[ \tan \theta = -|\tan \theta| \].

Evaluate Case 1

For \[ \tan \theta = |\tan \theta| \], the equation simplifies to \[ (\sin \theta + \cos \theta) \tan \theta = \tan \theta \]. Dividing both sides by \[ \tan \theta \] (noting \[ \tan \theta e 0 \]), we get \[ \sin \theta + \cos \theta = 1 \].

Solve \( \sin \theta + \cos \theta = 1 \)

Square both sides to get \[ (\sin \theta + \cos \theta)^2 = 1 \], which simplifies to \[ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 \]. Using \[ \sin^2 \theta + \cos^2 \theta = 1 \], we end up with \[ 1 + 2 \sin \theta \cos \theta = 1 \]. This further simplifies to \[ \sin 2\theta = 0 \], giving solutions \[ 2\theta = n\pi \], hence \[ \theta = \frac{n\pi}{2} \] for \[ n = 0, 1, 2, \ldots \]. Identify valid intervals yielding four solutions: \[ \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \].

Evaluate Case 2

For \[ \tan \theta = -|\tan \theta| \], the equation becomes \[ (\sin \theta + \cos \theta) \tan \theta = -\tan \theta \]. Dividing both sides by \[ \tan \theta \], we get \[ \sin \theta + \cos \theta = -1 \]. Repeating similar steps, square both sides to derive \[ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 \], yielding \[ \sin 2\theta = 0 \. \]

Combine Results

Both cases ultimately simplify to \[ \sin 2\theta = 0 \], giving \[ \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \] in the interval \[ [0, 2\pi] \]. There are four solutions.

Key concepts

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, especially useful in solving various types of equations in the IIT JEE preparation. These functions include \( \sin(\theta) \), \( \cos(\theta) \), \( \tan(\theta) \), \( \sec(\theta) \), \( \csc(\theta) \), and \( \cot(\theta) \). In the given problem, we mainly deal with \(\tan(\theta)\) and \(\sec(\theta)\). It's crucial to remember some key identities:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)
• \( \sec^2(\theta) = 1 + \tan^2(\theta) \)

Understanding these identities helps simplify expressions and solve equations systematically. For instance, knowing that \( \sec^2(\theta) - 1 = \tan^2(\theta) \) allows us to simplify complex expressions into more familiar forms. Always keep these identities in mind; they are tools to unlock tricky problems.

Equation Solving

Solving trigonometric equations requires a clear strategy and understanding of math principles. In our exercise, simplifying the equation was the first step. By transforming \( \sec^2(\theta) - 1 \) into \( \tan^2(\theta) \) and taking the square root, we managed to reduce the complexity.

Next, we handled the absolute value, considering \(\tan(\theta)\) not only in its positive form but also its negative form. This bifurcation into two cases helps us tackle both possibilities.

When solving equations, it's essential to:

• Simplify expressions using known identities or substitutions.
• Consider all possible values for variables, especially when dealing with absolute values or square roots.
• Verify your solutions in the original equation to ensure they fit within given constraints or intervals.

The ultimate goal is to isolate the variable (\(\theta\) in this scenario) and find all possible solutions within the given domain, which, for the IIT JEE, is commonly [0, 2π].

Always go step-by-step: simplify, solve, and verify.

IIT JEE Preparation

Preparing for the IIT JEE requires mastering a broad range of topics, including trigonometry. Practicing problems like this one enhances problem-solving skills and familiarizes you with the variety of question styles in the exam.

Key tips for IIT JEE trigonometric equations:

• Understand and memorize fundamental trigonometric identities.
• Practice a mix of straightforward and complex problems to build confidence.
• Learn to quickly identify opportunities to simplify problems using identities.
• Regularly review and practice solving equations in different domains.

Trigonometry appears in various sections of the IIT JEE, not just in pure mathematics but also in physics problems. Thus, a solid foundation in this area is crucial.

To excel, allocate consistent study time to trigonometry, review previous years' problems, and take timed practice tests. By doing so, you'll not only bolster your trigonometry skills but also improve your overall problem-solving speed and accuracy.