Answer:
[tex]\text{tan}^2(\theta)\cdot \text{cos}^2(\theta)=\text{sin}^2(\theta)[/tex]
Step-by-step explanation:
We have been given a trigonometric identity [tex]\text{tan}^2(\theta)\cdot \text{cos}^2(\theta)=1-\text{cos}^2(\theta)[/tex]. We are asked to determine the first step of the proof.
We will use identity [tex]\text{sin}^2(\theta)+\text{cos}^2(\theta)=1[/tex] to prove our given identity.
From above identity, we will get:
[tex]\text{sin}^2(\theta)+\text{cos}^2(\theta)-\text{cos}^2(\theta)=1-\text{cos}^2(\theta)[/tex]
[tex]\text{sin}^2(\theta)=1-\text{cos}^2(\theta)[/tex]
So, we will substitute [tex]\text{sin}^2(\theta)=1-\text{cos}^2(\theta)[/tex] is our given identity as:
[tex]\text{tan}^2(\theta)\cdot \text{cos}^2(\theta)=\text{sin}^2(\theta)[/tex]
Therefore, the first line of the proof would be [tex]\text{tan}^2(\theta)\cdot \text{cos}^2(\theta)=\text{sin}^2(\theta)[/tex].
Upon dividing both sides of equation by [tex]\text{cos}^2(\theta)[/tex], we will get:
[tex]\text{tan}^2(\theta)=\frac{\text{sin}^2(\theta)}{\text{cos}^2(\theta)}[/tex]
[tex]\text{tan}^2(\theta)=(\frac{\text{sin}(\theta)}{\text{cos}(\theta)}})^2[/tex]
[tex]\text{tan}^2(\theta)=(\text{tan}(\theta)})}^2[/tex]
[tex]\text{tan}^2(\theta)=\text{tan}^2(\theta)[/tex]
Hence proved.
