Question

Simplifying Trigonometric Expressions-

Write the trigonometric expression $\frac{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{\theta }\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}$in terms of sine and cosine, and then simplify.

Short answer

On simplifying expression we get $\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}$

Step-by-step solution

Step 1. Given information

A trigonometric expression as$\frac{\sec \theta -\cos \theta }{\sin \theta }$

Step 2. Concept used

To simplify the given expression, we need to use trigonometric identities. We will convert it into simplest form and that will be our answer.

Step 3. Calculation

Now, write $\frac{\sec \theta -\cos \theta }{\sin \theta }$in form of $sineand\cos ine$:

$\begin{array}{l}\frac{\sec \theta -\cos \theta }{\sin \theta }=\frac{\frac{1}{\cos \theta }-\cos \theta }{\sin \theta }\\ \left(\sec \theta =\frac{1}{\cos \theta }\right)\\ \frac{\sec \theta -\cos \theta }{\sin \theta }=\frac{\frac{1-{\cos }^2\theta }{\cos \theta }}{\sin \theta }\\ \frac{\sec \theta -\cos \theta }{\sin \theta }=\frac{1-{\cos }^2\theta }{\sin \theta \times \cos \theta }\\ \left(1-{\cos }^2\theta ={\sin }^2\theta \right)\\ \frac{\sec \theta -\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta }{\sin \theta \times \cos \theta }\\ \frac{\sec \theta -\cos \theta }{\sin \theta }=\frac{\sin \theta }{\cos \theta }\end{array}$