Question

I need help on a pre-calc problem: Verify the identity.

cosine of x divided by quantity one plus sine of x plus quantity one plus sine of x divided by cosine of x equals two times secant of x.

Answer 1

Here's a better view of it (cos x/ 1 + sin x) + (1 + sin x/ cos x) = 2 sec x

Answer 2

\[\Large\rm \frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}=2\sec x\]

Answer 3

I guess what I would do isssss...
Multiply the first fraction by the `conjugate` of the `denominator`.
That should get us on the right track.

Don't mess with the second fraction for now.

Answer 4

\[\Large\rm \color{royalblue}{\left(\frac{1-\sin x}{1-\sin x}\right)}\frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}\]

Answer 5

Remember what happens when you multiply conjugates? :)

Answer 6

What do you mean specifically?

Answer 7

(1-sin x) and (1+sin x) are called conjugates because they are `the same` except for the sign between them.

When multiplying conjugates, you end up with the difference of squares:\[\Large\rm (a-b)(a+b)=a^2-b^2\]

Answer 8

Oh, okay. So you'd end up with a denominator of 1+sinx^2?

Answer 9

\[\Large\rm\frac{(1-\sin x)\cos x}{1-\sin^2 x}+\frac{1+\sin x}{\cos x}\]Ok great!
Let's NOT do the multiplication in the top, just leave it like this for now.

Answer 10

Hmm, anything we can do with 1-sin^2x?
Do you remember your Pythagorean Identity involving sine and cosine?

Answer 11

sin^2x+cos^2x=1

Answer 12

Hmmm, yah.
Can we use that maybe?
If we subtract sin^2x from each side,\[\Large\rm \cos^2x=1-\sin^2x\]Ooo, how bout that?

Answer 13

Woah, okay this was what I was missing. Okay so next would we dot he same for the second fraction's numerator or?

Answer 14

No no. Let's leave the second fraction alone.
We're about to run into a common denominator almost by accident here.

Answer 15

\[\Large\rm\frac{(1-\sin x)\cos x}{\cos^2x}+\frac{1+\sin x}{\cos x}\]So we're here, yes?
In the first fraction, any common factors between the numerator and denominator that we can cancel out?

Answer 16

Oh so it'd be \[1-sinx/cosx\]

Answer 17

\[\Large\rm\frac{(1-\sin x)\cancel{\cos x}}{\cos^{\cancel{2}}x}+\frac{1+\sin x}{\cos x}\]
\[\Large\rm\frac{(1-\sin x)}{\cos x}+\frac{1+\sin x}{\cos x}\]Mmm yes, very good!

Answer 18

Wow, okay, now we have a common denominator so we can add right. and then could we use the sin and cos Pythagorean identity?

Answer 19

Correct, let's just simply add the tops from there.\[\Large\rm \frac{1-\sin x+1+\sin x}{\cos x}\]No, we won't need to use the Pythagorean Identity again.

Answer 20

Combine like-terms in the numerator c:
What are you left with on top?

Answer 21

just 2 right?

Answer 22

\[\Large\rm \frac{2}{\cos x}\]Ok good.
Let's write it like this, pulling the 2 in front as multiplication,\[\Large\rm 2\frac{1}{\cos x}\]Another identity should jump out at you ;)

Answer 23

Awh yes. now we have 2secx. Thank you so so so so much. My algebra skills just fly out the window when we involve identities so thank you.

Answer 24

Yayyy team \c:/
No problem.