prove (cot x)/(cosec x - 1) = (cosec x + 1)/(cot x)
thanks, i just cant seem to prove it and have tried a few methods but can't get it all the way through.
jacs
Expanding $\displaystyle \cot$ and $\displaystyle \csc$ into $\displaystyle \cos$s and $\displaystyle \sin$s the identity to prove becomes:Originally Posted by jacs
$\displaystyle
\frac{\cos(x)}{1-\sin(x)}=\frac{1+\sin(x)}{\cos(x)}
$
Now look at the right hand side of this:
$\displaystyle
\frac{1+\sin(x)}{\cos(x)}=\frac{1+\sin(x)}{\cos(x) }\times \frac{1-\sin(x)}{1-\sin(x)}=$$\displaystyle
\frac{1-\sin^2(x)}{\cos(x)(1-\sin(x))}=\frac{\cos^2(x)}{\cos(x)(1-\sin(x))}
$$\displaystyle
=\frac{\cos(x)}{1-\sin(x)}
$
Which proves the result.
RonL
Hello, jacs!
Here's another approach ... which probably didn't occur to you either.
. . It's clever, but rather obscure.
We're expected to know the identity: .$\displaystyle \csc^2\theta - 1\:=\:\cot^2\theta$
Prove: .$\displaystyle \frac{\cot x}{\csc x - 1} \:=\:\frac{\csc x + 1}{\cot x}$
On the left side, multiply top and bottom by $\displaystyle \csc x + 1:\;\;\frac{\cot x}{\csc x - 1}\cdot\frac{\csc + 1}{\csc x + 1}$
. . $\displaystyle =\;\frac{\cot x(\csc x + 1)}{\csc^2x - 1} \;= \;\frac{\cot x(\csc x + 1)}{\cot^2x} \;= \;\frac{\csc x + 1}{\cot x}$ . . . ta-DAA!
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This is one of a set of eight similar identities.
. . They all suggest "multiplying by the conjugate".
They look like this:
. . $\displaystyle \frac{\sin x}{1 + \cos x} \;=\;\frac{1 - \cos x}{\sin x}$ . . . and: $\displaystyle \frac{\sin x}{1 - \cos x}\;=\;\frac{1 + \cos x}{\sin x}$
. . $\displaystyle \frac{\cos x}{1 + \sin x} \:=\:\frac{1 - \sin x}{\cos x}$ . . . and: $\displaystyle \frac{\cos x}{1 - \sin x}\;=\;\frac{1 + \sin x}{\cos x}$
. . $\displaystyle \frac{\tan x}{\sec x + 1}\;=\;\frac{\sec x - 1}{\tan x}$ . . . and: $\displaystyle \frac{\tan x}{\sec x - 1}\;=\;\frac{\sec x + 1}{\tan x}$
. . $\displaystyle \frac{\cot x}{\csc x+ 1}\;=\;\frac{\csc x- 1}{\cot x}$ . . . and: $\displaystyle \frac{\cot x}{\csc x- 1} \;= \;\frac{\csc x + 1}{\cot x}$
Do we have to memorize them? . . . No!
When you see something that remotely resembles these forms,
. . immediately "cross multiply".
If we get a "Pythagorean identity", it is one of these forms.
For example, test the first equation: .$\displaystyle \frac{\sin x}{1 + \cos x}\:=\:\frac{1 - \cos x}{\sin x}$
"Cross-multiply": .$\displaystyle \sin x\cdot\sin x\;=\;(1+\cos x)(1 - \cos x)$
. . and we get: .$\displaystyle \sin^2x\;=\;1 - \cos^2x\quad\Rightarrow\quad \sin^2x + \cos^2x\:=\:1$ . . . see?
Since I'm self-teaching myself, one of the questions I had, when I was working on these trigonometric identities, was what is the purpose of shifting from one form to another? In exercise after exercise, the solution revolved around ginning up a multiplier (not always as simple as the conjugate, or shifting into one form or another of the pythagorean identities). To say this is frustrating is an understatement, and after working through dozens of these sorts of problems, I came to the conclusion that you had to be inside the head of the problem makers to know which way to go to arrive at the solution. Very annoying, but more importantly, here is my question. Is this sort of manipulation useful today, or is this a carryover from techniques that were advantageous when there wern't calculators? This is a serious question.
To be more exact, regarding my last comment. I'm not denigrating the trigonometric identities. The identities are fine and useful. I can even live with these little one step and two step mutations, such as illustrated in this thread. What I am referring to exactly are the 6 to 7 step monstrosities, twisted, contorted shiftings to another form that, on the face of it, looks as ugly (or pretty, if you prefer) as the original! And, adding to my suspicion that these doings are the remnant of the pre-calculator days, there were no applied problems in that section of my book...just endless mutations of equations to no obvious purpose. Enlighten me please as to how this is useful.
I would say probably not a lot of use, the only trig identies that I findOriginally Posted by spiritualfields
usefully are the sin/cos version of Pythagoras's theorem, and the formulae
for the sin and cos of the sum of angles. Most problems can be solved with
these and:
$\displaystyle e^{\bold{i}\theta}=\cos(\theta)+\bold{i} \sin(\theta)$
(In fact the stuff mentioned above can be derived from this in one-liners)
RonL