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- Thread starter chlobuggy
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- Thread starter
- #1

- Jan 26, 2012

- 890

We want to show that:1 + 1 = 2 sec^{2}x

______ ________

1+sinx1-sinx

PLEASE SOMEONE HELP!!

UGH! In case you can't tell what that says, it's 1/1+sinx+ 1/1-sinx= 2sec^{2}x

\[\frac{1}{1+\sin(x)}+\frac{1}{1-\sin(x)}=2 \sec^2(x)\]

Well:

\[\frac{1}{1+\sin(x)}+\frac{1}{1-\sin(x)}=\frac{(1-\sin(x))+(1-\sin(x))}{(1+\sin(x))(1-\sin(x))}=\frac{2}{1-\sin^2(x)}=\frac{2}{\cos^2(x)}\]

etc...

CB

- Jun 30, 2012

- 176

(With apologies to Captain Black: sorry, I didn't realize that you had undertaken to help this newcomer till I had laboriously put up my post, after which I felt disinclined to waste the whole thing by pulling it down: i certainly DID NOT want to seem like I was jumping in, and would not have done so, had I realized that you had already answered)1 + 1 = 2 sec^{2}x

______ ________

1+sinx1-sinx

PLEASE SOMEONE HELP!!

UGH! In case you can't tell what that says, it's 1/1+sinx+ 1/1-sinx= 2sec^{2}x

Hellow ChloBuggy and welcome to the site

First, you need to learn to use the resources of the site. Have a look around and learn how to write equations so that others will be able to better assist you.

Second, why the urgency?

In the meantime, what you have here is a simple trigonometric identity, based on the core identity of Pythagoras, which is:

$$cos^2(x)+sin^2(x)=1$$

It is the manipulation of this identity which will allow you to solve your equation, which I think must be as follows:

$$\frac{1}{1+sin(x)}+\frac{1}{1-sin(x)}=\frac{2}{1-sin^2(x)}$$

$1-sin^2(x)=cos^2(x)$ (Refer to the initial Pythagorean identity)

This gives you: $$\frac{2}{cos^2(x)}=sec^2(x)$$

Divide both sides by 2 and you are left with your identity:

$$sec^2(x)=sec^2(x)$$

However, unless you know why you are doing this, I am afraid what I have written will not help much.

But I offer it nonetheless in case it spurs you on to learn some more about the site and to get some more help from those better equipped to do so than I, who am a beginner at this.

Best regs,

DeusAbscondus