# Solving Identites

## Homework Statement

hey i really need some ones help i need to prove these 2 identities. im in grade 12 and ive been working on these questions for almost 2 hours and keep failing plzzz help
b. 1/1+sinx = sec2x - tanx/cosx

-secx=1/cosx
-tanx=sinx/cosx
-sin2x+cos2x=1
-cscx=1/sinx

## The Attempt at a Solution

So i tried proving the identity and got stuck...
i picked one side which was 1/1+sinx
then i got the conjugate
=1/1+sinx(1-sinx/1-sinx)
=1-sinx(1/sinx)
and this is where i have no idea what to do because i need to make it equal to sec2x-tanx/cosx

## Answers and Replies

cristo
Staff Emeritus
By sec2x do you mean sec2x? If so, the equation is $$\frac{1}{1+sinx}=sec^2x-\frac{tanx}{cosx}$$.

Now, my advice would be to write out tanx and secx in terms of cosx and sinx. Then multiply the equation to get rid of the denominators. Try this and post what you get.

Tom Mattson
Staff Emeritus
Gold Member
=1/1+sinx(1-sinx/1-sinx)

I take it that you mean:

$$\frac{1}{1+\sin(x)}\frac{1-\sin(x)}{1-\sin(x)}$$

=1-sinx(1/sinx)

I'm not sure of how you got this (or even of how to read it!), but it definitely looks wrong. You should multiply across the top and across the bottom in my expression above. Try that, and post what you get.

so i worked it out and im pretty sure its 1- sinx/ cosx

Tom Mattson
Staff Emeritus
Gold Member
Nope, the cosine should be squared.

no the equation is exacly what cristo got

ok so its 1-sinx over cos2x

Tom Mattson
Staff Emeritus
Gold Member
no the equation is exacly what cristo got

No, I'm talking about the part of your solution that I quoted.

wait no its 1/cos2x-sinx/cosx

Tom Mattson
Staff Emeritus
Gold Member
ok so its 1-sinx over cos2x

Do you mean cos2(x)? If so, then you're right. But if you mean cos(2x), then you're not right.

no i mean cos squared 2 x but i still havent solved the identity and i dont know where to go from here because ill just be making the equation more complicated by converting the identities

Tom Mattson
Staff Emeritus
That's wrong. When you multiply $1+\sin(x)$ by $1-\sin(x)$, you get $1-\sin^2(x)=\cos^2(x)$. The argument of the cosine is not $2x$.