Question

Simplify the trigonometric expression

Answer 1

\[1\div1+\sin \theta + 1\div 1 - \sin \theta \]

Answer 2

\[\frac{ 1 }{ 1+\sin \theta }+\frac{ 1 }{ 1-\sin \theta } \]

Well, you want to make these into one fraction by regular common denominator rules. So if you knowhow to properly combine those into one fraction, things will just kind of simplify together.

Answer 3

well im not asking for an answer but could you help by explaining step by step how to get the answer

Answer 4

Right. I wasn't sure if youd know how to go about making those into one fraction or not. Well, I want to multiply the two denominators together. Now just like when finding common denominators with all numbers, whatever I multiply into the bottom, I must multiply into the top. So it would kind of look like this:

\[\frac{ (1-\sin \theta)1+(1+\sin \theta)1 }{ (1+\sin \theta)(1-\sin \theta) }\]

I multiplied the two denominators together. The first fraction on the left was multiplied by 1-sinx on the bottom, so I multiplied it by that on the top. The right fraction was multiplied by 1+sinx on the bottom, so then it was multiplied by 1+sinx on the top. Doing this means they now have a common denominator and can be written as one fraction like above. Now the top can simplify pretty easily:

\[\frac{ 2 }{ (1+\sin \theta)(1-\sin \theta) }\]

Now you just need to multiply out the bottom. You see what I did up to this point?

Answer 5

okay!:) that helps! would the answer have (sin)? my answers have (cos) (sec) (csc) and (cot)

Answer 6

There would be no sin in your answer. After you multiply the bottom out you can use an identity

Answer 7

well i do kno the answer is 2(???)^2theta

Answer 8

but idk about the identity

Answer 9

Did you multiply the bottomout properly?

Answer 10

yes i did. but i dont understand the identity part

Answer 11

Can I see what you ended up with?

Answer 12

\(\bf \cfrac{ 1 }{ 1+sin(\theta) }+\cfrac{ 1 }{ 1-sin(\theta) } \implies \cfrac{(1-sin(\theta))+(1+sin(\theta))}{(1+sin(\theta))(1-sin(\theta))}\\
\cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{(1+sin(\theta))(1-sin(\theta))}
\implies \cfrac{2}{(1+sin(\theta))(1-sin(\theta))}\)

as already pointed out by Psymon

Answer 13

now if you keep in mind that "difference of squares" is => \(\bf (a-b)(a+b) = (a^2-b^2)\)
then we could say that \(\bf (1+sin(\theta))(1-sin(\theta)) \implies 1^2-sin^2(\theta) \implies 1-sin^2(\theta)\)

thus then \(\bf \cfrac{2}{(1+sin(\theta))(1-sin(\theta))} \implies \cfrac{2}{1-sin^2(\theta)}\\
\textit{recall the trig identity of } \\sin^2(\theta)+cos^2(\theta)=1 \implies cos^2(\theta)= 1-sin^2(\theta)\)

Answer 14

so, can you get it from there?