Answer:
See Below.
Step-by-step explanation:
We want to prove the trigonometric identity:
[tex]\displaystyle \frac{\sec^2(\theta)-1}{\sin(\theta)}=\frac{\sin(\theta)}{1-\sin^2(\theta)}[/tex]
To start, let's simplify the right side. Recall the Pythagorean Identity:
[tex]\sin^2(\theta)+\cos^2(\theta)=1[/tex]
Therefore:
[tex]\cos^2(\theta)=1-\sin^2(\theta)[/tex]
Substitute:
[tex]\displaystyle \frac{\sin(\theta)}{1-\sin^2(\theta)}=\frac{\sin(\theta)}{\cos^2(\theta)}[/tex]
Split:
[tex]\displaystyle =\frac{\sin(\theta)}{\cos(\theta)}\left(\frac{1}{\cos(\theta)}\right)=\tan(\theta)\sec(\theta)[/tex]
Therefore, our equation becomes:
[tex]\displaystyle \frac{\sec^2(\theta)-1}{\sin(\theta)}=\tan(\theta)\sec(\theta)[/tex]
From the Pythagorean Identity, we can divide both sides by cos²(θ). This yields:
[tex]\displaystyle \tan^2(\theta)+1=\sec^2(\theta)[/tex]
So:
[tex]\tan^2(\theta)=\sec^2(\theta)-1[/tex]
Substitute:
[tex]\displaystyle \frac{\tan^2(\theta)}{\sin(\theta)}=\tan(\theta)\sec(\theta)[/tex]
Rewrite:
[tex]\displaystyle (\tan(\theta))^2\left(\frac{1}{\sin(\theta)}\right)=\tan(\theta)\sec(\theta)[/tex]
Recall that tan(θ) = sin(θ)/cos(θ). So:
[tex]\displaystyle \frac{\sin^2(\theta)}{\cos^2(\theta)}\left(\frac{1}{\sin(\theta)}\right)=\tan(\theta)\sec(\theta)[/tex]
Simplify:
[tex]\displaystyle \frac{\sin(\theta)}{\cos^2(\theta)}=\tan(\theta)\sec(\theta)[/tex]
Simplify:
[tex]\tan(\theta)\sec(\theta)=\tan(\theta)\sec(\theta)}[/tex]
Hence proven.
