Question

Proving Identities

Verify the identity$\frac{\tan y}{\csc y}=\frac{1}{\cos y}-\frac{1}{\sec y}$

Short answer

The expression$\frac{\tan y}{\csc y}=\frac{1}{\cos y}-\frac{1}{\sec y}$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression as$\frac{\tan y}{\csc y}=\frac{1}{\cos y}-\frac{1}{\sec y}$

Step 2. Concept used.

Break it down into two parts: first, solve or simplify the LHS. Then simplify RHS and compare the two outcomes. If two replies are the same, it's an identity.

Step 3. Calculation.

Now, simplify LHS of $\frac{\tan y}{\csc y}=\frac{1}{\cos y}-\frac{1}{\sec y}$:

$$\begin{gathered} \frac{\tan y}{\csc y}=\frac{\frac{\sin y}{\cos y}}{\frac{1}{\sin y}} \\ =\frac{\sin y\times \sin y}{\cos y} \\ =\frac{{\sin }^{2}y}{\cos y} \end{gathered}$$

RHS of the equation is-

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

Both RHS and LHS are equal. Hence it is an identity.