# Solving Identites

ritagogna

## Homework Statement

hey i really need some ones help i need to prove these 2 identities. I am in grade 12 and I've been working on these questions for almost 2 hours and keep failing pleasezz 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

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.

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.

ritagogna
so i worked it out and I am pretty sure its 1- sinx/ cosx

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

ritagogna
no the equation is exacly what cristo got

ritagogna
ok so its 1-sinx over cos2x

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.

ritagogna
wait no its 1/cos2x-sinx/cosx

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.

ritagogna
no i mean cos squared 2 x but i still haven't solved the identity and i don't know where to go from here because ill just be making the equation more complicated by converting the identities

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 $2x$.