Answer:
See Below.
Step-by-step explanation:
Problem 1)
We want to verify that:
[tex]\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)[/tex]
Note that cot(x) = cos(x) / sin(x). Hence:
[tex]\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)[/tex]
Multiply:
[tex]\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)[/tex]
Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:
[tex]\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)[/tex]
Split:
[tex]\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)[/tex]
Simplify:
[tex]\csc(x)-\sin(x)=\csc(x)-\sin(x)[/tex]
Problem 2)
We want to verify that:
[tex]\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Square:
[tex]\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:
[tex]\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Factor out the sin²(x) from the denominator:
[tex]\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Factor (perfect square trinomial):
[tex]\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:
[tex]\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Factor (difference of two squares):
[tex]\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Factor out a negative from the first factor in the denominator:
[tex]\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Cancel:
[tex]\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}[/tex]
Distribute the negative into the numerator. Therefore:
[tex]\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}[/tex]
