# Identities!

• Oct 31st 2006, 06:16 PM
anna_sims
Identities!
I need some guidance on proving these identitles. Any help would be appreciated.

1)

((sec x+1)(sec x-1))/(2+(tan x+1)(tan x-1)) = sin^2 x

2)

sec x + csc x = (1+tan x/sin x)

:cool:
• Oct 31st 2006, 06:46 PM
ThePerfectHacker
Quote:

Originally Posted by anna_sims

2)

sec x + csc x = (1+tan x/sin x)

Work on the right side,
$\displaystyle \frac{1+\tan x}{\sin x}$

$\displaystyle \frac{1}{\sin x}+\frac{\tan x}{\sin x}$

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

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

$\displaystyle \csc x+\sec x$
• Oct 31st 2006, 06:51 PM
topsquark
Quote:

Originally Posted by anna_sims
1)

((sec x+1)(sec x-1))/(2+(tan x+1)(tan x-1)) = sin^2 x

Well first of all:
$\displaystyle sin^2x + cos^2x = 1$

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

So
$\displaystyle tan^2x + 1 = sec^2x$

Now to the LHS of your expression:
$\displaystyle \frac{(secx + 1)(secx - 1)}{2 + (tanx + 1)(tanx - 1)}$ Mulitplying out the numerator and denominator:

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

$\displaystyle \frac{tan^2x}{tan^2 + 1}$ <-- Using the tan^2 relation in the numerator.

$\displaystyle \frac{tan^2x}{sec^2x}$ <-- Using the tan^2 relation in the denominator.

Now convert everything into sines and cosines:
$\displaystyle \frac{ \frac{sin^2x}{cos^2x} }{ \frac{1}{cos^2x} }$

Multiply top and bottom by $\displaystyle cos^2x$:
$\displaystyle \frac{ \frac{sin^2x}{cos^2x} }{ \frac{1}{cos^2x} } \cdot \frac{cos^2x}{cos^2x}$ = $\displaystyle \frac{sin^2x}{1} = sin^2x$

-Dan