Problem 19 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

After simplification, both the LHS and RHS are \(\frac{1}{\sin t \cos t}\), thus verifying the original identity \(\frac{\csc ^{2} t}{\cot t}=\csc t \sec t\).

Step 1: Break down LHS into terms of basic trigonometric functions

Begin by writing the left-hand side (LHS) of the equation in terms of basic trigonometric functions sine and cosine. \(\csc t\) is the reciprocal of \(\sin t\) and \(\cot t\) is the reciprocal of \(\tan t\), or in other terms \(\frac{\cos t}{\sin t}\), therefore, the left-hand-side of the equation can be written as: \(\frac{1}{\sin^2 t}\times\frac{\sin t}{\cos t}\)

Step 2: Simplify trigonometric equation

Simplify the equation from step 1, by canceling out the common factors. In the multiplication, the \(\sin t\) in the numerator and denominator will cancel each other out, resulting in: \(\frac{1}{\sin t \cos t}\)

Step 3: Convert RHS into terms of basic trigonometric functions

Now, handle the right-hand side (RHS) of the equation similarly to step 1. \(\csc t\) represents the reciprocal of \(\sin t\) and \(\sec t\) is the reciprocal of \(\cos t\), that gives us \(\frac{1}{\sin t}\times\frac{1}{\cos t} = \frac{1}{\sin t \cos t}\)

Step 4: Compare LHS and RHS

Notice that after simplification, both sides of the equation are equal. Thus the original identity is verified: \(\frac{\csc ^{2} t}{\cot t}=\csc t \sec t\).