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

Given the function:\[f(\theta) = \frac{\tan(\tan(\theta)) - \tan(\sin(\theta))}{\tan(\theta) - \sin(\theta)}.\]The function is continuous at \( \theta = 0 \), implying \( \lim_{\theta \to 0} f(\theta) = f(0) \).To determine \( f(0) \), we evaluate the limit of \( f(\theta) \) as \( \theta \) approaches 0.Step 1: Taylor Series ExpansionFor small \( \theta \), the Taylor series expansions are:- \( \tan(\theta) = \theta + \frac{\theta^3}{3} + O(\theta^5) \)- \( \sin(\theta) = \theta - \frac{\theta^3}{6} + O(\theta^5) \)Step 2: Approximate Numerator and DenominatorUsing these expansions:- \( \tan(\tan(\theta)) \approx \theta + \frac{\theta^3}{3} \)- \( \tan(\sin(\theta)) \approx \theta - \frac{\theta^3}{6} \)Numerator:\[\tan(\tan(\theta)) - \tan(\sin(\theta)) \approx \left(\theta + \frac{\theta^3}{3}\right) - \left(\theta - \frac{\theta^3}{6}\right) = \frac{\theta^3}{2}.\]Denominator:\[\tan(\theta) - \sin(\theta) \approx \left(\theta + \frac{\theta^3}{3}\right) - \left(\theta - \frac{\theta^3}{6}\right) = \frac{\theta^3}{2}.\]Step 3: Simplify ExpressionSubstituting approximations into \( f(\theta) \):\[f(\theta) \approx \frac{\frac{\theta^3}{2}}{\frac{\theta^3}{2}} = 1.\]Therefore, \( f(0) = 1 \).