Question

If \( f(\theta) = \frac{\tan(\tan(\theta)) - \tan(\sin(\theta))}{\tan(\theta) - \sin(\theta)} \) is continuous at \( \theta = 0 \), then the value of \( f(\theta) \) at \( \theta = 0 \) is equal to.

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When dealing with limits of functions that involve trigonometric expressions, expand them using Taylor series for small angles to simplify the evaluation.

Answer 1

The Correct Option is B

We are given the function: \[ f(\theta) = \frac{\tan(\tan(\theta)) - \tan(\sin(\theta))}{\tan(\theta) - \sin(\theta)}. \] The function is continuous at \( \theta = 0 \), meaning that \( \lim_{\theta \to 0} f(\theta) = f(0) \). To find the value of \( f(\theta) \) at \( \theta = 0 \), we need to evaluate the limit of the given expression as \( \theta \) approaches 0. We can start by checking the behavior of the numerator and denominator as \( \theta \to 0 \). ### Step 1: Use Taylor series expansion For small values of \( \theta \), we can expand the functions using their Taylor series around \( \theta = 0 \): - \( \tan(\theta) = \theta + \frac{\theta^3}{3} + O(\theta^5) \) - \( \sin(\theta) = \theta - \frac{\theta^3}{6} + O(\theta^5) \) ### Step 2: Approximate the numerator and denominator Using the above expansions: - \( \tan(\tan(\theta)) = \tan(\theta + \frac{\theta^3}{3}) \approx \theta + \frac{\theta^3}{3} \) - \( \tan(\sin(\theta)) = \tan(\theta - \frac{\theta^3}{6}) \approx \theta - \frac{\theta^3}{6} \) Thus, the numerator becomes: \[ \tan(\tan(\theta)) - \tan(\sin(\theta)) \approx \left(\theta + \frac{\theta^3}{3}\right) - \left(\theta - \frac{\theta^3}{6}\right) = \frac{\theta^3}{2}. \] The denominator becomes: \[ \tan(\theta) - \sin(\theta) \approx \left(\theta + \frac{\theta^3}{3}\right) - \left(\theta - \frac{\theta^3}{6}\right) = \frac{\theta^3}{2}. \] ### Step 3: Simplify the expression Now, we substitute these approximations into the original expression for \( f(\theta) \): \[ f(\theta) \approx \frac{\frac{\theta^3}{2}}{\frac{\theta^3}{2}} = 1. \] Thus, the value of \( f(\theta) \) at \( \theta = 0 \) is 1.