Answer:
Proofs are in the explanation.
Step-by-step explanation:
b) My first thought is to divide top and bottom on the left hand side by [tex]\cos(\alpha)[/tex].
I see this would give me 1 on top and where that sine is, it would give me tangent since sine/cosine=tangent.
Let's do it and see:
[tex]\frac{\cos(\alpha)}{\cos(\alpha)-\sin(\alpha)} \cdot \frac{\frac{1}{\cos(\alpha)}}{\frac{1}{\cos(\alpha)}}[/tex]
[tex]=\frac{\frac{\cos(\alpha)}{\cos(\alpha)}}{\frac{\cos(\alpha)}{\cos(\alpha)}-\frac{\sin(\alpha)}{\cos(\alpha)}}[/tex]
[tex]=\frac{1}{1-\tan(\alpha)}[/tex]
c) My first idea here is to expand the cos(x+y) using the sum identity for cosine.
So let's do that:
[tex]\frac{\cos(x)\cos(y)-\sin(x)\sin(y)}{\cos(x)\sin(y)}[/tex]
Separating the fraction:
[tex]\frac{\cos(x)\cos(y)}{\cos(x)\sin(y)}-\frac{\sin(x)\sin(y)}{\cos(x)\sin(y)}[/tex]
The cos(x) cancel's in the first fraction and the sin(y) cancels in the second fraction:
[tex]\frac{\cos(y)}{\sin(y)}-\frac{\sin(x)}{\cos(x)}[/tex]
[tex]\cot(y)-\tan(x)[/tex]
d) This one makes me think it is definitely essential that we use properties of logarithms.
The left hand side can be condense into one logarithm using the product law:
[tex]\ln|(1+\cos(\theta))(1-\cos(\theta))|[/tex]
We are multiplying conjugates inside that natural log so we only need to multiply the first and the last:
[tex]\ln|1-\cos^2(\theta)|[/tex]
I can rewrite [tex]1-\cos^2(\theta)[/tex] using the Pythagorean Identity:
[tex]\sin^2(\theta)+\cos^2(\theta)=1[/tex]:
[tex]\ln|\sin^2(\theta)|[/tex]
Now by power rule for logarithms:
[tex]2\ln|\sin(\theta)|[/tex]