Problem 48 - Precalculus Essentials

Question

Verify each identity. $$\frac{\csc t-1}{\cot t}=\frac{\cot t}{\csc t+1}$$

Step-by-Step Solution

Answer

By following these steps, it can be observed that \( \frac{\csc t-1}{\cot t} = \frac{\cot t}{\csc t+1} \) is indeed a valid identity as starting from the LHS and going through the steps we end up with the RHS

Step 1: Rewrite csc and cot as reciprocal identities

The first step is to rewrite all trigonometric functions in terms of sine and cosine. This will make it easier to simplify the expression. So, rewrite the expression as follows: \[ \frac{1/\sin t - 1}{\cos t / \sin t} = \frac{\cos t / \sin t}{1/\sin t + 1} \]

Step 2: Simplify each side separately

Now, simplifying each side, we get \[ \frac{\sin t - 1}{\cos t} = \frac{\cos t }{\sin t + 1} \]

Step 3: Multiplication by conjugate

Multiplying each side of the equation by the conjugate of the denominator to get rid of the fractions. This means multiplying the LHS of the equation by (\(\sin t + 1\)) and the RHS by \((\cos t + 1)\) we get: \[ \frac{\sin t - 1}{\cos t} \cdot (\sin t + 1) = \frac{\cos t}{\sin t + 1} \cdot (\cos t + 1) \] Upon simplifying we get, \[ \sin^2 t - 1 = \cos^2 t - 1\]

Step 4: Use Pythagorean identity

By using the Pythagorean identity, \(\sin^2 t + \cos^2 t = 1\), we can replace \(\sin^2 t\) with \(1 - \cos^2 t\) and \(\cos^2 t\) with \(1 - \sin^2 t\). This leads us to: \[1 - \cos^2 t - 1 = 1 -\sin^2 t - 1\] which simplifies to \( \cos^2 t = \sin^2 t \)