Chapter 14.4, Problem 10CYU

To determine

To Prove: The value $\tan \text{θ}=\frac{1-\cos 2\text{θ}}{\sin 2\text{θ}}$ is an identity.

Answer to Problem 10CYU

  $\tan \text{θ}=\frac{1-\cos 2\text{θ}}{\sin 2\text{θ}}$ is an identity.

Explanation of Solution

Given information:

The given expression is $\tan \text{θ}=\frac{1-\cos 2\text{θ}}{\sin 2\text{θ}}$ .

Formula used:

  $\begin{array}{l}1-\cos 2\text{θ}=\text{2}{\sin }^2\text{θ}\\ \sin 2\text{θ}=\text{2sinθcosθ}\end{array}$

Calculation:

  $\begin{array}{l}\Rightarrow \tan \text{θ}=\frac{1-\cos 2\text{θ}}{\sin 2\text{θ}}\\ \Rightarrow \tan \text{θ}=\frac{2{\sin }^2\text{θ}}{2\sin \text{θcosθ}}\\ \Rightarrow \tan \text{θ}=\frac{\sin \text{θ}}{\cos \text{θ}}\\ \Rightarrow \tan \text{θ}=\tan \text{θ}\\ \therefore \tan \text{θ}=\frac{1-\cos 2\text{θ}}{\sin 2\text{θ}}\text{is}\text{an}\text{identity}\text{.}\end{array}$