• Feb 16th 2011, 05:47 PM
TacticalPro
Hi I have a quiz tomorrow and have been looking over some review problems that will be similar to those on the quiz. I've stumbled across a hard one and really need help. I've attempted numerous approaches and still haven't figured it out, so can someone walk me through?

THANKS! (i removed x's for visual purposes)

((tan^2)-1)/(sec^2) = (tan-cot)/(tan+cot)
• Feb 16th 2011, 06:22 PM
topsquark
Quote:

Originally Posted by TacticalPro
Hi I have a quiz tomorrow and have been looking over some review problems that will be similar to those on the quiz. I've stumbled across a hard one and really need help. I've attempted numerous approaches and still haven't figured it out, so can someone walk me through?

THANKS! (i removed x's for visual purposes)

((tan^2)-1)/(sec^2) = (tan-cot)/(tan+cot)

As usual, I have a way, but I doubt it's the most direct. Take a look at the RHS. Write everything in terms of sines and cosines, add the fractions in the numerator and denominator, then simplify. I get that
$\displaystyle \frac{tan(x) - cot(x)}{tan(x) + cot(x)} = sin^2(x) - cos^2(x)$

Now multiply the top and bottom of this expression by $\displaystyle sec^2(x) = \frac{1}{cos^2(x)}$. Simplify this a bit and you get your LHS.

-Dan
• Feb 16th 2011, 06:22 PM
alexmahone
Quote:

Originally Posted by TacticalPro
Hi I have a quiz tomorrow and have been looking over some review problems that will be similar to those on the quiz. I've stumbled across a hard one and really need help. I've attempted numerous approaches and still haven't figured it out, so can someone walk me through?

THANKS! (i removed x's for visual purposes)

((tan^2)-1)/(sec^2) = (tan-cot)/(tan+cot)

Substitute $sec^2x=tan^2x+1$.
• Feb 16th 2011, 07:14 PM
TacticalPro
Quote:

Originally Posted by topsquark
As usual, I have a way, but I doubt it's the most direct. Take a look at the RHS. Write everything in terms of sines and cosines, add the fractions in the numerator and denominator, then simplify. I get that
$\displaystyle \frac{tan(x) - cot(x)}{tan(x) + cot(x)} = sin^2(x) - cos^2(x)$

Now multiply the top and bottom of this expression by $\displaystyle sec^2(x) = \frac{1}{cos^2(x)}$. Simplify this a bit and you get your LHS.

-Dan

I don't know you get sin^2 - cos^2

i get (sin^2+cos^2)/(sin^2-cos^2)
• Feb 16th 2011, 08:35 PM
topsquark
Quote:

Originally Posted by TacticalPro
I don't know you get sin^2 - cos^2

i get (sin^2+cos^2)/(sin^2-cos^2)

$\displaystyle \frac{tan(x) - cot(x)}{tan(x) + cot(x)} = \frac{ \frac{sin(x)}{cos(x)} - \frac{cos(x)}{sin(x)} }{ \frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)} }$

Now clear the complex fractions:
$\displaystyle = \frac{ \frac{sin(x)}{cos(x)} - \frac{cos(x)}{sin(x)} }{ \frac{sin(x)}{cos(x)} + \frac{cos(x)}{sin(x)} } \cdot \frac{sin(x)~cos(x)}{sin(x)~cos(x)} = \frac{sin^2(x) - cos^2(x)}{sin^2(x) + cos^2(x)}$

and, of course, the denominator is 1.

Now multiply by
$\displaystyle \frac{ \frac{1}{cos^2(x)} }{ \frac{1}{cos^2(x)} }$

-Dan
• Feb 16th 2011, 08:51 PM
use $\sec^2x=\tan^2x+1$
$\displaystyle{\frac{\tan^2x-1}{\tan^2x+1}}$
divide up and down by $\tan x$