Find the derivative of $\frac {\sin\theta}{\cos\theta}$

Analysis

The question asks for the derivative of the trigonometric expression $\frac{\sin\theta}{\cos\theta}$ with respect to $\theta$.

Concept

1. Trigonometric Identity: Recall that $\frac{\sin\theta}{\cos\theta} = \tan\theta$.
2. Differentiation Rule: The derivative of $\tan\theta$ with respect to $\theta$ is a standard derivative in calculus: $$\frac{d}{d\theta}(\tan\theta) = \sec^2\theta$$
3. Alternative Method (Quotient Rule): If you prefer to use the quotient rule $\frac{d}{d\theta}(\frac{u}{v}) = \frac{v\frac{du}{d\theta} - u\frac{dv}{d\theta}}{v^2}$: • Let $u = \sin\theta \implies \frac{du}{d\theta} = \cos\theta$ • Let $v = \cos\theta \implies \frac{dv}{d\theta} = -\sin\theta$ • Applying the formula: $$\frac{(\cos\theta)(\cos\theta) - (\sin\theta)(-\sin\theta)}{\cos^2\theta}$$ $$\frac{\cos^2\theta + \sin^2\theta}{\cos^2\theta}$$ • Since $\cos^2\theta + \sin^2\theta = 1$ (Pythagorean identity): $$\frac{1}{\cos^2\theta} = \sec^2\theta$$

Solution

Both the direct derivative of the identity and the quotient rule lead to the same result:

Comparing this to the provided options: A. $\sec^2\theta$ B. $\tan\theta \csc\theta$ C. $\csc\theta \sec\theta$ D. $\csc^2\theta$

The result matches option A.

Tip

Always simplify trigonometric expressions before differentiating if possible. Converting $\frac{\sin\theta}{\cos\theta}$ to $\tan\theta$ first saves you the time and potential errors associated with the quotient rule!