OpenStudy (anonymous):

simplify the expression: 1+ sinx/cosx + cosx/1+ sinx do i multiply by the opposite denominator?

5 years ago
OpenStudy (anonymous):

5 years ago
OpenStudy (anonymous):

@Kitsune ty ^^ <3

5 years ago
OpenStudy (lgbasallote):

$\large \mathbf{\frac{1+sinx}{cosx} + \frac{cosx}{1+sinx}}$ ^that's the problem right?

5 years ago
OpenStudy (anonymous):

yup

5 years ago
OpenStudy (lgbasallote):

hmmm..your problems are getting interesting >:)) gimme a sec..

5 years ago
OpenStudy (anonymous):

okie dokie lol

5 years ago
OpenStudy (anonymous):

Try multiplying the second term by the conjugate of the denominator

5 years ago
OpenStudy (anonymous):

oooo i have it...give me a sec

5 years ago
OpenStudy (experimentx):

2/cosx <--- most likely you will get this answer.

5 years ago
OpenStudy (anonymous):

$\frac{1+\sin(x)}{\cos(x)} + \frac{\cos(x)}{1+\sin(x)}$ $\frac{[1+\sin(x)]^{2}}{\cos(x)[1+\sin(x)]} + \frac{[\cos(x)]^{2}}{\cos(x)[1+\sin(x)]}$ $\frac{1+2\sin(x) + \sin ^{2}(x) + \cos ^{2}(x)}{\cos(x)[1+\sin(x)]}$ $\frac{2+2\sin(x)}{\cos(x)[1+\sin(x)]}$ $\frac{2[1+\sin(x)]}{\cos(x)[1+\sin(x)]}$ $\frac{2}{\cos(x)}$

5 years ago
OpenStudy (anonymous):

the answer supposed to be 2secx o.o i just checked my answer sheet

5 years ago
OpenStudy (anonymous):

$\sec(x) = \frac{1}{\cos(x)}$ so 2/cos(x) = 2 * (1/cos(x)) or 2*sec(x) :P

5 years ago