Question

Trig help check my work and help with one problem! Attached

Answer 1

@jdoe0001

Answer 2

hmmm what's \(\bf \cfrac{1}{cos(x)}-\cfrac{cos(x)}{1} ?\)

Answer 3

seems like \(\bf \cfrac{cos^2(x)}{cos(x)}\).... not sure

Answer 4

secx-cosx

Answer 5

wel.... right.... that'd be the expanded version.... but the subtraction?

Answer 6

I have the whole thing wrong?

Answer 7

hmm well... is not very clear I don't think... there.... what \(\bf \cfrac{1}{cos(x)}-\cfrac{cos(x)}{1}\) seems to be suggestive of \(\bf \cfrac{1}{cos(x)}-\cfrac{cos(x)}{1}\implies
\cfrac{cos^2(x)}{cos(x)}\)

Answer 8

I am just not good at this haha

Answer 9

well lemme do a run down on it

Answer 10

thanks!

Answer 11

\(\bf \cfrac{sec(\theta)-cos(\theta)}{cos(\theta)}\implies
\cfrac{
\frac{1}{cos(x)}-\frac{cos(x)}{1}
}{
\frac{cos(x)}{1}
}\implies \cfrac{\frac{1\cdot 1-cos(\theta)\cdot cos(\theta)}{cos(x)}}{\frac{cos(x)}{1}}
\\ \quad \\
\cfrac{\frac{1-cos^2(\theta)}{cos(x)}}{\frac{cos(x)}{1}}\implies \cfrac{1-cos^2(\theta)}{cos(x)}\cdot \cfrac{1}{cos(x)}
\\ \quad \\
{\color{brown}{sin^2(\theta)+cos^2(\theta)=1\to sin^2(\theta)=1-cos^2(\theta)}}\qquad thus
\\ \quad \\
\cfrac{{\color{brown}{ 1-cos^2(\theta)}}}{cos(x)}\cdot \cfrac{1}{cos(x)}\implies \cfrac{{\color{brown}{ sin^2(\theta)}}}{cos(x)}\cdot \cfrac{1}{cos(x)}\implies \cfrac{sin^2(\theta)}{cos^2(\theta)}\)

Answer 12

I was way off