This has been alluding me for a week and I am now down to the wire

Prove that (sin^2(x)) / (1-cos(x)) equals (sec(x)+1)/(sec(x))

any help,hints,clues or outright answers would be appreciated!

Thanks,

mj

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- May 19th 2008, 04:50 PMmaryjaneproving trig identity
This has been alluding me for a week and I am now down to the wire

Prove that (sin^2(x)) / (1-cos(x)) equals (sec(x)+1)/(sec(x))

any help,hints,clues or outright answers would be appreciated!

Thanks,

mj - May 19th 2008, 04:54 PMJhevon
Consider the $\displaystyle LHS$

$\displaystyle \frac {\sin^2 x}{1 - \cos x} = \frac {1 - \cos^2 x}{1 - \cos x}$ ...........Pythagorean identity

$\displaystyle = \frac {(1 + \cos x )( 1 - \cos x)}{1 - \cos x}$ ..........Numerator was the difference of two squares

$\displaystyle = 1 + \cos x$ ..........................cancel

$\displaystyle = \frac {\sec x}{\sec x} + \frac 1{\sec x}$ .....................algebraic manipulation

$\displaystyle = \frac {\sec x + 1}{\sec x}$ ...........................combine the fractions

$\displaystyle = RHS$

Thus, $\displaystyle LHS \equiv RHS$, as desired - May 19th 2008, 04:54 PMMathstud28
$\displaystyle \frac{\sin^2(x)}{1-\cos(x)}=\frac{\sec(x)+1}{\sec(x)}$

we know that $\displaystyle \sin^2(x)=1-\cos^2(x)$

and we also know that $\displaystyle a^2-b^2=(a+b)(a-b)$

so $\displaystyle \sin^2(x)=1-\cos^2(x)=(1+\cos(x))(1-\cos(x))$

So

now subbing we get

$\displaystyle \frac{(1+\cos(x))(1-\cos(x))}{1-\cos(x)}=\frac{\sec(x)+1}{\sec(x)}$

Cancelling the left side and splitting the left we get

$\displaystyle 1+\cos(x)=\frac{\sec(x)}{\sec(x)}+\frac{1}{\sec(x) }$

now we know that $\displaystyle \frac{\sec(x)}{\sec(x)}=1$

and $\displaystyle \sec(x)=\frac{1}{\cos(x)}$

so $\displaystyle \frac{1}{\sec(x)}=\frac{1}{\frac{1}{\cos(x)}}=\cos (x)$

imputting back in gives

$\displaystyle 1+\cos(x)=1+\cos(x)$ - May 19th 2008, 05:03 PMmaryjane
well, well, well.... I now feel like an idiot

(Bow) thank-you both!

mj - May 19th 2008, 05:08 PMJhevon
- May 19th 2008, 05:33 PMmaryjane
(Wink) haha

one more question I have...how do we get

__sec(x)__+__1__

sec(x) sec(x)

from the 1 + cos(x) ... I realize we can sub the first part for the 1

but how do we sub__1__

sec(x) from cos(x) ?

I am obviously missing something (i wish i could find it)

mj - May 19th 2008, 05:49 PMMathstud28