Question

complete following identity: 1/1+sinx + 1/1-sinx
I'm so lost on this question.

Answer 1

try adding
\[\frac{1}{1+b}+\frac{1}{1-b}\]
first to see what you get. then replace b by sine

Answer 2

i will write it if you like

Answer 3

any additional info would be appreciated as this blows my mind.

Answer 4

what else would you do here?

Answer 5

ok lets do the algebra first. usually these work like this
1) some algebra
2) a trig identity
3) some more algebra (maybe)
so rather than cluttering up your paper with sines lets just add

Answer 6

\[\frac{1}{1+b}+\frac{1}{1-b}=\frac{1-b+1+b}{(1+b)(1-b)}\]

Answer 7

simplify to get
\[\frac{2}{1-b^2}\]notice this part had nothing at all to do with trig

Answer 8

now replace b by sine and write
\[\frac{2}{1-\sin^2(x)}\] and make the one trig step here, namely recall that
\[\cos^2(x)+\sin^2(x)=1\] so
\[1-\sin^2(x)=\cos^2(x)\]

Answer 9

you get
\[\frac{2}{\cos^2(x)}=2\sec^2(x)\]

Answer 10

thank you so much. this makes more sense than the tutors do for class.