Answer:
[tex]\displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1[/tex]
Step-by-step explanation:
Prove that:
[tex]\displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1[/tex]
Recall that by definition:
[tex]\displaystyle \tan x=\frac{\sin x}{\cos x}[/tex]
Therefore,
[tex]\displaystyle \tan^2x=\left (\frac{\sin^2x}{\cos^2x}\right)^2=\frac{\sin^2x}{\cos^2x}[/tex]
Substitute [tex]\displaystyle \tan^2x=\frac{\sin^2x}{\cos^2x}[/tex] into [tex]\displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1[/tex]:
[tex]\displaystyle \frac{1}{\sin^2x}-\frac{1}{\frac{\sin^2x}{\cos^2x}}=1[/tex]
Simplify:
[tex]\displaystyle \frac{1}{\sin^2x}-\frac{\cos^2x}{\sin^2x}=1[/tex]
Combine like terms:
[tex]\displaystyle \frac{1-\cos^2x}{\sin^2x}=1[/tex]
Recall the following Pythagorean Identity:
[tex]\sin^2x+\cos^2x=1[/tex] (derived from the Pythagorean Theorem)
Subtract [tex]\cos^2x[/tex] from both sides:
[tex]\sin^2=1-\cos^2x[/tex]
Finish by substituting [tex]\sin^2=1-\cos^2x[/tex] into [tex]\displaystyle \frac{1-\cos^2x}{\sin^2x}=1[/tex]:
[tex]\displaystyle \frac{\sin^2x}{\sin^2x}=1,\\\\1=1\:\boxed{\checkmark\text{ True}}[/tex]
