# Trig Identity Problems

• Nov 28th 2008, 08:51 PM
Chouman
Trig Identity Problems
Got two problems for extra credit that I can't seem to figure out if anyone here can.

Prove the following.

1. cosx-(sinx-1)/cosx+(sinx-1) = sinx/1-cosx It says as a hint to multiply
the numerator and
denominator of the left side
by cosx-(sinx-1)

2. cotx/1-tanx + tanx/1-cotx - 1 = secxcscx

If anyone happens to know how to do these, please let me know, thanks.
• Nov 28th 2008, 11:42 PM
xeno_vert
Quote:

Originally Posted by Chouman
Got two problems for extra credit that I can't seem to figure out if anyone here can.

Prove the following.

1. cosx-(sinx-1)/cosx+(sinx-1) + sinx/1-cosx It says as a hint to multiply
the numerator and
denominator of the left side
by cosx-(sinx-1)

2. cotx/1-tanx + tanx/1-cotx - 1 = secxcscx

If anyone happens to know how to do these, please let me know, thanks.

I already solved the number 1 but i can't solve the number 2...

Here's the solution:

cosx - (sinx - 1) / cosx = (sinx - 1) = sinx / 1 - cosx

*you'll cancel -(sinx - 1) and + (sinx -1) since that the first term is expressed in negative while the other is positive (outside the parentheses)

cosx / cos x = sinx / 1- cosx

1 = sinx / sinx

1=1

i hope it'll help...
• Nov 29th 2008, 08:51 AM
Soroban
Hello, Chouman!

Quote:

$1)\;\text{Prove: }\;\frac{\cos x-\sin x +1}{\cos x+ \sin x-1} \:=\:\frac{\sin x}{1-\cos x}$ . . Ignore their hint

Multiply top and bottom by $(\cos x + \sin x - 1)$

. . $\frac{\cos x - \sin x + 1}{\cos x + \sin x - 1}\cdot {\color{blue}\frac{\cos x + \sin x - 1}{\cos x + \sin x - 1}}$

Numerator: . $\cos^2\!x + {\color{red}\rlap{/////////}}\sin x\cos x - {\color{green}\rlap{/////}}\cos x - {\color{red}\rlap{/////////}}\sin x\cos x - \sin^2x + \sin x + {\color{green}\rlap{/////}}\cos x + \sin x - 1$

. . . $= \;2\sin x - \sin^2\!x - (1 - \cos^2\!x)$

. . . $=\;2\sin x - \sin^2\!x - \sin^2\!x$

. . . $=\;2\sin x - 2\sin^2\!x$

. . . $= \;2\sin x(1 - \sin x)$

Denominator: . $\cos^2\!x + \sin x\cos x - \cos x + \sin x\cos x + \sin^2\!x - \sin x - \cos x - \sin x + 1$

. . . $= \;\underbrace{\cos^2\!x + \sin^2\!x}_{\text{This is 1}} + 1 - 2\sin x - 2\cos x + 2\sin x\cos x$

. . . $= \;2 - 2\sin x - 2\cos x + 2\sin x\cos x$

. . . $= \;2(1-\sin x) - 2\cos x(1 - \sin x)$

. . . $= \;2(1-\sin x)(1 - \cos x)$

And the fraction becomes: . $\frac{{\color{green}\rlap{/}}2\cdot\sin x\cdot({\color{red}\rlap{////////}}1 - \sin x)}{{\color{green}\rlap{/}}2\cdot({\color{red}\rlap{////////}}1-\sin x)\cdot(1 - \cos x)} \;=\;\frac{\sin x}{1 - \cos x} \quad\hdots$ ta-DAA!

• Nov 29th 2008, 09:42 AM
Soroban
Hello again, Chouman!

I thinik the second one is even worse ... (Is that possible?)

Quote:

$2)\;\;\frac{\cot x}{1-\tan x} + \frac{\tan x}{1-\cot x} - 1 \;=\; \sec x\csc x$

The left side is: . $\frac{\dfrac{\cos x}{\sin x}} {1 - \dfrac{\sin x}{\cos x}} + \frac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}} - 1$

Multiply top and bottom by $\sin x\cos x$ ... both fractions.

. . $\frac{\sin x\cos x\left(\dfrac{\cos x}{\sin x}\right)} {\sin x\cos x\left(1 - \dfrac{\sin x}{\cos x}\right)} + \frac{\sin x\cos x\left(\dfrac{\sin x}{\cos x}\right)}{\sin x\cos x\left(1-\dfrac{\cos x}{\sin x}\right)} \;-\; 1$

. . $=\;\frac{\cos^2\!x}{\sin x\cos x - \sin^2\!x} + \frac{\sin^2\!x}{\sin x\cos x - \cos^2\!x} \:- \:1$

. . $= \;\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} + \frac{\sin^2\!x}{\cos x(\sin x - \cos x)} \:-\:1$

. . $= \;\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} \:{\color{red}-}\: \frac{\sin^2\!x}{\cos x(\cos x \:{\color{red}-}\: \sin x)} \:=\:1$

. . $= \;{\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} - {\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^2\!x}{\cos x(\cos x - \sin x)} \:-\:1$

. . $= \;\frac{\overbrace{\cos^3\!x - \sin^3\!x}^{\text{diff. of cubes}}}{\sin x\cos x(\cos x -\sin x)} \:-\:1$

. . $= \;\frac{({\color{red}\rlap{///////////}}\cos x - \sin x)(\cos^2\!x + \sin x\cos x + \sin^2\!x)}{\sin x\cos x({\color{red}\rlap{///////////}}\cos x - \sin x)}\:-\:1$

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

. . $= \;\frac{1}{\sin x\cos x} + \frac{\sin x\cos x}{\sin x\cos x} \:-\: 1 \quad=\quad \frac{1}{\sin x\cos x} \:+\: 1 \:-\: 1$

. . $=\;\frac{1}{\cos x\sin x} \quad=\quad \sec x\csc x \quad\hdots$ Whew!

• Nov 29th 2008, 10:26 AM
Chouman
Thanks a lot both of you. I appreciate it. The 2nd question most likely is harder because my teacher said they usually get harder as they progress and the 2nd question was after the 1st one.