Problem 70 - Precalculus Essentials

Question

Rewrite each expression in terms of the given function or functions. $$\frac{1}{\sin x \cos x}-\cot x ; \cot x$$

Step-by-Step Solution

Answer

\(\cot^2 x - \cot x\)

Step 1: Simplify the First Term

The first term \( \frac{1}{\sin x \cos x} \) can be rewritten using the reciprocal of sine and cosine, which gives \( \frac{1} {\sin x} \times \frac{1} {\cos x} \). Because \( \frac{1} {\sin x} \) equals \( \csc x \) and \( \frac{1} {\cos x} \) equals \( \sec x \), the first term can be simplified as \( \csc x \cdot \sec x \). Substituting cotangent for cosecant and secant, we obtain \( \cot x \cdot \cot x \) or \( \cot^2 x \)

Step 2: Rewrite the Second Term

The second term is already \( \cot x \) and does not need to be rewritten.

Step 3: Combine Both Terms

Combine the simplified first term \( \cot^2 x \) and the second term \( - \cot x \) to get the final answer.