Problem 22 - Precalculus Essentials

Question

Verify each identity. $$\frac{\cot ^{2} t}{\csc t}=\csc t-\sin t$$

Step-by-Step Solution

Answer

The short answer to the exercise is that the presented equation is indeed valid as both sides simplify to the same expression, justifying the identity.

Step 1: Rewrite in terms of sines and cosines

Write the given equation in terms of sines and cosines. Because \(\cot t=\frac{\cos t}{\sin t}\) and \(\csc t=\frac{1}{\sin t}\), it is possible to rewrite the left-hand side of the equation as \(\frac{(\frac{\cos t}{\sin t})^{2}}{(\frac{1}{\sin t})}\). Similarly \(\csc t - \sin t\) can be rewritten as \(\frac{1}{\sin t} - \sin t \).

Step 2: Simplify the Equation

Further simplifying the equation results in \(\frac{\cos^{2} t}{\sin^{2} t} = \frac{1}{\sin t} - \sin t\). This can be manipulated to form \(\frac{\cos^{2} t}{\sin^{2} t} = \frac{1 - \sin^{2} t}{\sin t} \).

Step 3: Use Pythagorean Identity

Using the Pythagorean identity, which states that \(\cos^{2}t + \sin^{2}t = 1\), allows for simplification of the equation. Substitute \(1 - \sin^{2} t\) for \(\cos^{2} t\) in the equation to get \(\frac{1 - \sin^{2} t}{\sin^{2} t} = \frac{1 - \sin^{2} t}{\sin t}\).