Question

verifying an identity

(cos x/1+sin x ) + (1+sinx/cosx) = 2 sec x

So I started on the left and ended up with 2 + 2 sin x/ 1+ 2 sinx + cos x. I'm not sure what to do next? Can anyone spot me where I'm wrong?

Answer 1

\[\large \frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}\]Hmm I'm not really sure what you did to get that result... hmmm

Answer 2

Oh ok I think I see where you got that :3 Doing something like this I guess?\[\large \frac{\cos^2 x}{\cos x(1+\sin x)}+\frac{(1+\sin x)^2}{\cos x(1+\sin x)}\]

Answer 3

Here is what got for the top cos^2 +1+2sin x + sin 2 x than 1 - sin^2x + 1 + 2 sin x + 2 sin ^ 2 x

Answer 4

Yes I Am ^

Answer 5

Hmm I think the approach we want to take is a little different.
We're going to get a common denominator using a different method, it will just kind of, show up.

\[\large \frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}\]Let's multiply the left term (top and bottom) by the conjugate of the bottom.\[\large \color{royalblue}{\frac{1-\sin x}{1-\sin x}}\cdot \frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}\]

Answer 6

\[\large \frac{\cos x(1-\sin x)}{1-\sin^2 x}+\frac{1+\sin x}{\cos x}\]Can you see the next step?
Hint ~ Important identity involving \(\large 1-\sin^2x\).

Answer 7

Hmm I'm guessing 1-sin^2x Cancels out on the right side

Answer 8

Recall the first identity they teach you in Trig.\[\large \cos^2x+\sin^2x=1\]From this identity we can derive this,\[\large 1-\sin^2x=\cos^2x\]So we're going to use this identity to replace our \(\large 1-\sin^2x\) with \(\large \cos^2 x\).

Answer 9

Oh! Thats right, what occurs next? Excuse my slowness I'm having a tough time with these identities

Answer 10

\[\large \frac{\cos x(1-\sin x)}{\color{royalblue}{1-\sin^2 x}}+\frac{1+\sin x}{\cos x} \qquad \rightarrow \qquad \frac{\cos x(1-\sin x)}{\color{royalblue}{\cos^2x}}+\frac{1+\sin x}{\cos x}\]
We'll cancel out a cosine from the top and bottom of the left term.\[\large \frac{\cancel{\cos x}(1-\sin x)}{\cos^{\cancel21}x}+\frac{1+\sin x}{\cos x} \qquad \rightarrow \qquad \frac{(1-\sin x)}{\cos x}+\frac{1+\sin x}{\cos x}\]
How bout that! We were able to get a common denominator by doing something rather sneaky :D

Answer 11

no way you could do that?! There are hidden rules in these identities... I'm guessing the top sin would cancel out and you would be left with 1/ cos x^2?

Answer 12

oopsie cos^x

Answer 13

Yes the sine's on top will cancel, but the 1's will add together, giving us,\[\large \frac{2}{\cos x}\]
Yes good catch, no square on the bottom. Just adding.

Answer 14

alright.. hmm so that would translate to 2 sec x? according to the properties?

Answer 15

\[\large \frac{(1-\sin x)}{\cos x}+\frac{1+\sin x}{\cos x} \qquad \rightarrow \qquad \frac{1-\sin x+1+\sin x}{\cos x}\]

Answer 16

\[\large \frac{2}{\cos x} \qquad = \qquad 2 \cdot \frac{1}{\cos x}\]Then we have to remember another important identity.\[\large \sec x=\frac{1}{\cos x}\]

Answer 17

the 2 comes in front am i right? :3

Answer 18

Yes :o

Answer 19

Thank you very much ! :DDD

Answer 20

c: