Question

Someone please help me with trig

Answer 1

\[\frac{ cosx }{ 1+sinx } + \frac{ 1+sinx }{ cosx } = 2secx\]

Answer 2

add

Answer 3

i need a step by step process

Answer 4

\[\frac{a}{1+b}+\frac{1+b}{a}=\frac{a^2+(1+b)^2}{a(1+b)}\] it has nothing to do with trig as a first step, it is how you add fractions

Answer 5

ok and then how do i do the problem then?

Answer 6

replace \(a\) by \(\cos(x)\) and \(b\) by \(\sin(x)\)

Answer 7

ok so \[\frac{ cosx }{ 1 + sinx } + \frac{ 1+sinx }{ cosx } = \frac{ \cos ^{2}+(1+\sin)^2 }{ \cos(1+\sin) }\]

Answer 8

right

Answer 9

then square in the numerator, giving
\[\frac{\cos^2(x)+1+2\sin(x)+\sin^2(x)}{\cos(x)(1+\sin(x))}\]

Answer 10

now comes the only real trig part
in the numerator you have \(\sin^2(x)+\cos^2(x)\) which is \(1\)

Answer 11

than get
\[\frac{2+2\sin(x)}{\cos(x)(1+\sin(x))}\] factor out the 2 up top, then cancel to get the answer you want

Answer 12

how do i factor it out?

Answer 13

never mind

Answer 14

someone please help

Answer 15

you factor out the two like this
\[2+2\sin(x)=2\left(1+\sin(x)\right)\]

Answer 16

then cancel the \(1+\sin(x)\) top and bottom

Answer 17

ok and then your left with 2/cos ?

Answer 18

yes

Answer 19

which equals 2secx right?