OpenStudy (lovelyharmonics):

Verify the identity.

OpenStudy (lovelyharmonics):

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.

OpenStudy (isaiah.feynman):

Use the equation editor please!

OpenStudy (lovelyharmonics):

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

OpenStudy (isaiah.feynman):

Too bad I don't have too much time to go step by step. But add the fractions on the left side and simplify carefully.

OpenStudy (lovelyharmonics):

i need someone to explain HOW to do this

OpenStudy (lovelyharmonics):

@Luigi0210

OpenStudy (anonymous):

Boobs

OpenStudy (luigi0210):

Get common denominators and then add thm

OpenStudy (lovelyharmonics):

SERIOUSLY flutterING STOP

OpenStudy (anonymous):

I love making her angry.

OpenStudy (lovelyharmonics):

i dont know how to get the denominators though though

OpenStudy (lovelyharmonics):

@Luigi0210

OpenStudy (anonymous):

I only got a 90% on the whole trig identities test so.

OpenStudy (lovelyharmonics):

.-.

OpenStudy (anonymous):

;-;

OpenStudy (luigi0210):

Work with it as if it was a regular fraction..

OpenStudy (anonymous):

I'll work with you like a regular fraction. >:D

OpenStudy (luigi0210):

*sigh* Okay, we have \[\frac{ \cos x }{ 1+\sin x } + \frac{ 1+\sin x }{ \cos x } =2\sec x\] Get common denominators: \[\frac{ \cos x }{ 1+\sin x }*(\frac{cosx}{cosx}) + \frac{ 1+\sin x }{ \cos x }*(\frac{1+sinx}{1+sinx}) =2\sec x\] Simifying it \[\frac{ \cos^2 x }{ 1+\sin x(cosx) } + \frac{ (1+\sin x)^2 }{(1+sinx) \cos x } =2\sec x\] Now simplify the tops \[\frac{cos^2x}{(1+sinx)(cosx)}+\frac{(1+2sinx+sin^2x)}{(1+sinx)(cosx)}\] Combine into one. \[\frac{sin^2x+cos^2x+2sinx+1}{(1+sinx)(cosx)}\] Use the trig identity that states \(sin^2x+cos^2x=1\) and sub So now we have \[\frac{1+2sinx+1}{(1+sinx)(cosx)}\] The top simplifies into \(2sinx+2\) which can be factored into \(2(1+sinx))\) So the \((1+sinx)\) cancel each other and we are left with \[\frac{2}{cosx}\] Which is \[large 2secx\] And that proves your identity.. @halorazer no, just no..

OpenStudy (luigi0210):

*Which is \[\large 2secx\]

OpenStudy (anonymous):

no worries brotaco, am straight.