# proving trig identity

• May 19th 2008, 04:50 PM
maryjane
proving 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 PM
Jhevon
Quote:

Originally Posted by maryjane
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

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 PM
Mathstud28
Quote:

Originally Posted by maryjane
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

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

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 PM
maryjane
well, well, well.... I now feel like an idiot
(Bow) thank-you both!

mj
• May 19th 2008, 05:08 PM
Jhevon
Quote:

Originally Posted by maryjane
well, well, well.... I now feel like an idiot
(Bow) thank-you both!

mj

*spider-man voice*Mary Jane! hang On! :D

you're welcome (Sun) ..........that's Kriz's icon :D
• May 19th 2008, 05:33 PM
maryjane
(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 PM
Mathstud28
Quote:

Originally Posted by maryjane
(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

If you are takling about how $\displaystyle \frac{\sec(x)+1}{\sec(x)}=1+\cos(x)$

its because $\displaystyle \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

the rest I showed in my repsonse