To verify the trigonometric identity
[tex]\[
\sec \theta - \cos \theta = \tan \theta \sin \theta,
\][/tex]
we will simplify both sides of the equation and show that they are indeed equal.
Firstly, recall the definitions of the trigonometric functions involved:
[tex]\[
\sec \theta = \frac{1}{\cos \theta},
\][/tex]
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}.
\][/tex]
### Left-Hand Side (LHS)
Consider the left-hand side of the identity:
[tex]\[
\sec \theta - \cos \theta.
\][/tex]
Substitute \(\sec \theta\) with \(\frac{1}{\cos \theta}\):
[tex]\[
\frac{1}{\cos \theta} - \cos \theta.
\][/tex]
To combine the terms, we need a common denominator:
[tex]\[
\frac{1 - \cos^2 \theta}{\cos \theta}.
\][/tex]
Using the Pythagorean identity, \(1 - \cos^2 \theta = \sin^2 \theta\), we rewrite the expression as:
[tex]\[
\frac{\sin^2 \theta}{\cos \theta}.
\][/tex]
### Right-Hand Side (RHS)
Now, consider the right-hand side of the identity:
[tex]\[
\tan \theta \sin \theta.
\][/tex]
Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), substitute it into the expression:
[tex]\[
\left(\frac{\sin \theta}{\cos \theta}\right) \sin \theta.
\][/tex]
This simplifies to:
[tex]\[
\frac{\sin^2 \theta}{\cos \theta}.
\][/tex]
### Compare Both Sides
We have:
[tex]\[
\text{LHS} = \frac{\sin^2 \theta}{\cos \theta}
\][/tex]
and
[tex]\[
\text{RHS} = \frac{\sin^2 \theta}{\cos \theta}.
\][/tex]
Since both sides simplify to the same expression, we have verified that:
[tex]\[
\sec \theta - \cos \theta = \tan \theta \sin \theta.
\][/tex]
Thus, the identity is verified.
