# Prove that each of the following is an identity

• Apr 18th 2009, 01:44 PM
starlet19
Prove that each of the following is an identity
I am frustated with these two problems, i can not figure them out, ive tried many identities but still no correct answer. (Thinking)

The problems are
#1.) (1+cot(x))^2-cot(x)=1/(1-cos(x))(1+cos(x))
#2.) 1-2cos^2(y)/1-2cos(y)sin(y)=sin(y)+cos(y)/sin(y)-cos(y)

• Apr 18th 2009, 02:32 PM
Calculus26
For # 2 multiply the top and bottom by sin(y) -cos(y)

expand and use sin^2(y) +cos^2(y) =1

For # 1 I'm not sure it is an identity since the right hand side

is csc^2(x) = cot^2(x) + 1 and if you expand the left hand side you get

cot^2(x) +2cot(x) +1 -cot(x) = cot^2(x) +cot(x) +1
• Apr 18th 2009, 02:39 PM
Soroban
Hello, starlet19!

The first one is not an identity . . . There must be a typo.

Quote:

$\displaystyle 2)\;\;\frac{1-2\cos^2\!y}{1-2\cos y\sin y} \:=\:\frac{\sin y+\cos y}{\sin y-\cos y}$
Multiply the right side by $\displaystyle \frac{\sin y - \cos y}{\sin y - \cos y}$

$\displaystyle \frac{\sin y + \cos y}{\sin y - \cos y}\cdot\frac{\sin y - \cos y}{\sin y - \cos y} \;=\;\frac{\sin^2\!y - \cos^2\!y}{\sin^2\!y - 2\cos y\sin y + \cos^2\!y}$

. . $\displaystyle =\;\frac{(1-\cos^2\!y) - \cos^2\!y}{\underbrace{\sin^2\!y + \cos^2\!y}_{\text{This is 1}}\: -\: 2\cos y\sin y} \;=\;\frac{1-2\cos^2\!y}{1 - 2\cos y\sin y} \quad\hdots\quad\text{ta-}DAA!$

Ahhh, too fast for me, Calculus26!
.