Trigonometry proof trouble

[SherryNugil]

On my homework, my prof asked us to prove the following eqivalency:

tan(x) - cos(x) = (sin(x)/cot(x)) - (1/cot(x)) - sec(x)

I have tried an seemingly endless number of transformations trying to make the two sides show an obvious equivalency without any luck.

I used the following identies in my attempts:
tan(x) = (sin(x)/cos(x))
cot(x) = (cos(x)/sin(x))
cot(x) = (1/tan(x))
sec(x) = (1/cos(x))

Also, do (cos(x))^2 = cos^2(x) ; (sin(x))^2 = sin^2(x) ; (tan(x))^2 = tan^2(x)?

After a while of getting nowhere, I decided to type each side of the original problem into my calculator to confirm that they were, in fact, equal. My result was that they are NOT equal and thus I cannot prove the equivalency as requested. Although, my results her could be wrong if I entered the expressions into my calculator incorrectly.

If anyone could provide me with some hints or explanations I would really appreciate it, as I have been stuck on this for a long time and feel like I must be overlooking something very obvious.

Thanks!

 

[royhaas]

Start with \(\displaystyle \tan (x) - \cos (x) = \sin (x) / \cos (x) - \cos (x)= \frac {\sin (x) -\cos^2 (x)}{\cos (x)}\)

 

[Unco]

G'day, Sherry.

On my homework, my prof asked us to prove the following eqivalency:

tan(x) - cos(x) = (sin(x)/cot(x)) - (1/cot(x)) - sec(x)
Are you sure it isn't:
tan(x) - cos(x) = (sin(x)/cot(x)) + (1/cot(x)) - sec(x) ?

If that's the case, although there are always many ways to do these, you might want to start with the more complicated-looking side, in this case the RHS, and simplify down to Royhaas's result of the LHS.

. . \(\displaystyle \L \frac{\sin{x}}{\cot{x}} \, = \, \frac{ \, \sin{x} \, }{\left(\frac{ \, \cos{x} \, }{ \, \sin{x}} \, \right)} \, = \, \frac{\sin^2{x}}{\cos{x}}\)

. . \(\displaystyle \L \frac{1}{\cot{x}} = \frac{\sin{x}}{\cos{x}}\)

. . \(\displaystyle \L -\sec{x} = -\frac{1}{\cos{x}}\)

Notice the common denominator of each term. (And don't forget \(\displaystyle \sin^2{x} \, + \, \cos^2{x} \, = \, 1\).)

 

[soroban]

Hello, Sherry!

I believe Unco found the typo . . . then the statement is true.

Prove: \(\displaystyle \,\tan(x)\,- \,\cos(x)\;=\;\frac{\sin(x)}{\cot(x)}\,\)+\(\displaystyle \,\frac{1}{\cot(x)}\,-\,\sec(x)\)

Also, do: \(\displaystyle (\cos(x))^2\,=\,\cos^2(x),\;\;(\sin(x))^2\,=\,\sin^2(x),\;\;(\tan(x))^2\,=\,\tan^2(x)\;\;\)yes!

The right side is: \(\displaystyle \,\sin(x)\cdot\tan(x)\,+\,\tan(x)\,-\,\sec(x)\)

\(\displaystyle \L\;\;\;=\;\sin(x)\cdot\frac{\sin(x)}{\cos(x)}\,+\,\frac{\sin(x)}{\cos(x)}\,-\,\frac{1}{\cos(x)} \;= \;\frac{\sin^2(x)\,+\,\sin(x)\,-\,1}{\cos(x)}\)

\(\displaystyle \L\;\;\;=\;\frac{\sin(x)\,-\,[1\,-\,\sin^2(x)]}{\cos(x)} \;= \;\frac{\sin(x)\,-\,\cos^2(x)}{\cos(x)}\;=\;\frac{\sin(x)}{\cos(x)}\,-\,\frac{\cos^2(x)}{\cos(x)}\)

\(\displaystyle \;\;\;=\;\tan(x)\,-\,\cos(x)\)

 

[SherryNugil]

Yes, thank you so much, that has got to be the typo. I can work through it now without the difficulty I was having before.

I also contacted my prof, so he will get back to me with the final word on how the equivalency should be changed. Thanks so much for your help! It was driving me mad.

You guys rule.