Question

Verify the identity.

Answer 1

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 2

Use the equation editor please!

Answer 3

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

Answer 4

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.

Answer 5

i need someone to explain HOW to do this

Answer 6

@Luigi0210

Answer 7

Boobs

Answer 8

Get common denominators and then add thm

Answer 9

SERIOUSLY flutterING STOP

Answer 10

I love making her angry.

Answer 11

i dont know how to get the denominators though though

Answer 12

@Luigi0210

Answer 13

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

Answer 14

.-.

Answer 15

;-;

Answer 16

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

Answer 17

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

Answer 18

*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..

Answer 19

*Which is \[\large 2secx\]

Answer 20

no worries brotaco, am straight.