help verifying identities

**adam456006** · March 15th 2009, 02:28 PM

are each of these an identity? show how...

#1

#2

#3

**e^(i*pi)** · March 15th 2009, 02:33 PM

Originally Posted by adam456006

are each of these an identity? show how...

#1

#2

#3

1. Use the difference of two squares:
$<br /> \frac{sin(u)(1+cos(u)}{(1-cos(u))(1+cos(u)} = \frac{(sin(u)(1+cos(u))}{1-cos^2(u)}$

$1-cos^2(u) = sin^2(u)$

$\frac{1+cos(u)}{sin(u)}$ so that's an identity

**adam456006** · March 15th 2009, 03:01 PM

I was working with # 2 and I don't think it is an identity, but I'm not sure....

**Soroban** · March 15th 2009, 03:08 PM

Hello, adam456006!

Are each of these an identity?

$(1)\;\;\frac{\sin u}{1-\cos u} - \frac{1+\cos u}{\sin u } \:=\:0$

Get a common denominator and subtract . . .

$\frac{\sin u}{1-\cos u} \cdot{\color{blue}\frac{\sin u}{\sin u}} - \frac{1+\cos u}{\sin u}\cdot{\color{blue}\frac{1-\cos u}{1-\cos u}} \;= \;\frac{\sin^2\! u - \overbrace{(1-\cos^2\!u)}^{\text{This is }\sin^2\!u}}{\sin u(1-\cos u)}$

. . $= \;\frac{\sin^2\!u-\sin^2\!u}{\sin u(1-\cos u)} \;=\;\frac{0}{\sin u(1-\cos u)} \;=\;0$

It is an identity.

$(2)\;\;\frac{\csc^2\!x - \cot^2\!x}{\sec x} \:=\:\sin x$

We have: . $\frac{\overbrace{\csc^2\!x - \cot^2\!x}^{\text{This is 1}}}{\sec x} \;=\;\frac{1}{\sec x} \;=\;\cos x$

It is not an identity.

$(3)\;\;\frac{\cos x - \cos y}{\sin x + \sin y} + \frac{\sin x - \sin y}{\cos x + \cos y} \:=\:0$

Get a common denominator and add . . .

$\frac{\cos x - \cos y}{\sin x + \sin y}\cdot{\color{blue}\frac{\cos x + \cos y}{\cos x + \cos y}} + \frac{\sin x - \sin y}{\cos x + \cos y}\cdot{\color{blue}\frac{\sin x + \sin y}{\sin x + \sin y}}$

. . $= \;\frac{\cos^2\!x - \cos^2\!y + \sin^2\!x - \sin^2\!y}{(\sin x+\sin y)(\cos x + \cos y)} \;=\;\frac{\overbrace{(\sin^2\!x +\cos^2\!x)}^{\text{This is 1}} - \overbrace{(\sin^2\!y + \cos^2\!y)}^{\text{This is 1}}}{(\sin x + \sin y)(\cos x + \cos y)}$

. . $= \;\frac{1-1}{(\sin x+\sin y)(\cos x + \cos y)} \;=\;\frac{0}{(\sin x+\sin y)(\cos x+\cos y)} \;=\;0$

It is an identity.

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