# verifying,simplifying, factoring trigs

• Apr 26th 2009, 07:53 PM
dmak263
verifying,simplifying, factoring trigs
I think I've tried every way possible, except the correct way.
Write the identity algebraically
1) (1+csc x)/(cot x + cos x)=sec x

factor the expression and use the fundamental identities to simplify
2)sec^4 x - tan^4 x

rewrite the expression so that it is not in fractional form.
3)tan^2 x/csc x + 1
• Apr 26th 2009, 09:55 PM
Prove It
Quote:

Originally Posted by dmak263
I think I've tried every way possible, except the correct way.
Write the identity algebraically
1) (1+csc x)/(cot x + cos x)=sec x

factor the expression and use the fundamental identities to simplify
2)sec^4 x - tan^4 x

rewrite the expression so that it is not in fractional form.
3)tan^2 x/csc x + 1

1. $\displaystyle \frac{1 + \csc{x}}{\cot{x} + \cos{x}} = \frac{1 + \frac{1}{\sin{x}}}{\frac{\cos{x}}{\sin{x}} + \cos{x}}$

$\displaystyle = \frac{\frac{\sin{x} + 1}{\sin{x}}}{\frac{\cos{x} + \cos{x}\sin{x}}{\sin{x}}}$

$\displaystyle = \frac{\sin{x} + 1}{\sin{x}} \times \frac{\sin{x}}{\cos{x} + \cos{x}\sin{x}}$

$\displaystyle = \frac{\sin{x} + 1}{\cos{x} + \cos{x}\sin{x}}$

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

$\displaystyle = \frac{1}{\cos{x}}$

$\displaystyle = \sec{x}$.
• Apr 26th 2009, 10:02 PM
Prove It
Quote:

Originally Posted by dmak263
I think I've tried every way possible, except the correct way.
Write the identity algebraically
1) (1+csc x)/(cot x + cos x)=sec x

factor the expression and use the fundamental identities to simplify
2)sec^4 x - tan^4 x

rewrite the expression so that it is not in fractional form.
3)tan^2 x/csc x + 1

2. $\displaystyle \sec^4{x} - \tan^4{x} = (\sec^2{x})^2 - (\tan^2{x})^2$

$\displaystyle = (\sec^2{x} + \tan^2{x})(\sec^2{x} - \tan^2{x})$

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

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

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

$\displaystyle = \frac{1 + \sin^2{x}}{\cos^2{x}}$

$\displaystyle = \frac{1 + 1 - \cos^2{x}}{\cos^2{x}}$

$\displaystyle = \frac{2 - \cos^2{x}}{\cos^2{x}}$

$\displaystyle = \frac{2}{\cos^2{x}} - \frac{\cos^2{x}}{\cos^2{x}}$

$\displaystyle = 2\sec^2{x} - 1$.
• Apr 26th 2009, 10:45 PM
Prove It
Quote:

Originally Posted by dmak263
I think I've tried every way possible, except the correct way.
Write the identity algebraically
1) (1+csc x)/(cot x + cos x)=sec x

factor the expression and use the fundamental identities to simplify
2)sec^4 x - tan^4 x

rewrite the expression so that it is not in fractional form.
3)tan^2 x/csc x + 1

3. Is this $\displaystyle \frac{\tan^2{x}}{\csc{x} + 1}$ or $\displaystyle \frac{\tan^2{x}}{\csc{x}} + 1$?
• Apr 27th 2009, 09:04 AM
dmak263
for #3, it's the first equation.
• Apr 27th 2009, 10:04 AM
masters
Quote:

Originally Posted by Prove It
3. Is this $\displaystyle \frac{\tan^2{x}}{\csc{x} + 1}$ or $\displaystyle \frac{\tan^2{x}}{\csc{x}} + 1$?

Quote:

Originally Posted by dmak263
for #3, it's the first equation.

Hi dmak263,

$\displaystyle \frac{\tan^2{x}}{\csc{x} + 1}$

Multiply top and bottom by $\displaystyle \csc x - 1$, and recall that $\displaystyle \csc^2 x-1=\cot^2 x$

$\displaystyle \frac{\tan^2 x(\csc x - 1)}{\csc^2 x - 1}=\frac{\tan^2 x(\csc x - 1)}{\cot^2 x}=\frac{\tan^2 x(\csc x - 1)}{\frac{1}{\tan^2 x}}=\tan^4 x(\csc x-1)$
• Apr 27th 2009, 11:11 AM
dmak263
I'm lost for words realizing how easy that was. Didn't see it at all. Thank you.