Question

How do you do this. I need help

cosx + sinx tanx = secx

Answer 1

\(\bf cos(x)+sin(x)+tan(x)\implies cos(x)+sin(x)+\cfrac{sin(x)}{cos(x)}\) add them up, see what you get

Answer 2

hmmm

Answer 3

is just hmmm a one sec

Answer 4

\(\bf cos(x)+sin(x)tan(x)\implies cos(x)+sin(x)\cfrac{sin(x)}{cos(x)}
\\ \quad \\
cos(x)+\cfrac{sin^2(x)}{cos(x)}\) there

is just a matter of summing them up
keeping in mind that \(\bf {\color{brown}{ sin^2(\theta)+cos^2(\theta)=1}}\)

Answer 5

is it 1/cosx + sin^2/cosx then. . . . yeah i really dont know. You keep on saying sin2(θ)+cos2(θ)=1 but i dont know what the use of it is for this problem.

Answer 6

hmm \(\bf cos(x)+\cfrac{sin^2(x)}{cos(x)}\implies \cfrac{cos(x)}{1}+\cfrac{sin^2(x)}{cos(x)}\) what would that give you?

Answer 7

cosx sin^2x/cosx . . . .

Answer 8

hmm well... is a sum though.... you're multiplying it there

so, if you sum them up... you'd see the numerator :)

Answer 9

So would you cancel the cosx and get sec^2x by itself. but what can i do to get secx by itself....

Answer 10

hmmm you have a sum of fractions
not a multiplicatoin of fractions

\(\bf cos(x)+\cfrac{sin^2(x)}{cos(x)}\implies \cfrac{cos(x)}{1}+\cfrac{sin^2(x)}{cos(x)}\ne \cfrac{cos(x)sin^2(x)}{cos(x)}
\)

for the sum, just like any other fraction sum, you use the LCD, to get the numerators

Answer 11

So would it be sin^2x/cosx = tanx?

Answer 12

well... lemme put it differently
\(\bf a+\cfrac{b}{a}\implies \cfrac{a}{1}+\cfrac{b}{a}\implies \cfrac{\square ?}{\square ?}\)
what would that give you?

Answer 13

ab/1a?

Answer 14

i take that back b/1

Answer 15

you may want to brush up fractions sum I'd think first -> http://www.youtube.com/watch?v=hw8aSrlJ-UQ

Answer 16

b/1 would be true if it were a * b/a
but is a + b/a

Answer 17

I just need to understand what you need to do after that. I dont think it would help watching how to sum fractions. I just suck at identities.

Answer 18

I know how to do fractions btw, i just dont understand the concept about identities. . .