Problem 20 - Precalculus Essentials

Question

Verify each identity. $$\frac{\sec ^{2} t}{\tan t}=\sec t \csc t$$

Step-by-Step Solution

Answer

So, we find that the original given identity \(\frac{\sec ^{2} t}{\tan t}=\sec t \csc t\) is true.

Step 1: Rewriting the Left-Hand Side

Rewrite the left-hand side (LHS) in terms of sine and cosine. The secant function can be written as the reciprocal of the cosine function and the tangent function can be written as sine divided by cosine. So, we get: \( \frac{1}{\cos^2t} \times \frac{\sin t}{\cos t}=\)\( \frac{\sin t}{\cos^3t} \)

Step 2: Simplifying the Left-Hand Side

We can simplify this equation further by separating it into two fractions like so: \(\frac{\sin t}{\cos t}\times \frac{1}{\cos^2t}=\) \(\frac{\sin t}{\cos t} \times \cos^{-2}t \)

Step 3: Rewriting in Terms of Secant and Cosecant

Now, rewrite \(\frac{\sin t}{\cos t}= \sec t\) which is just the reciprocal of the cosine function and \(\cos^{-2}t = \csc^{2}t\) which is the reciprocal of the sine function squared. So we have: \( \sec t \times \csc^{2}t \)