prove that 1/(sec(x)-tan(x)) = csc(x)-sin(x)

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jwpaine

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Using trig identities, prove that 1/(sec(x)-tan(x)) = csc(x)-sin(x)

What are good tricks to proving one side is equal to the other? I start converting either side to one of its identities, and start to get no where in matching to the other side.

I'll start with the right hand side:

I got 1/Sin(x) - Sin(x)

=

(1-Sin^2(x))/(Sin(x))

=

(cos^2(x))/(sin(x))

=

\(\displaystyle \L \frac{\frac{1}{sec^{2}x}}{sin(x)}\)

=

\(\displaystyle \L \frac{1}{sin(x) \cdot sec^{2}x}\)


And now I seem more lost than before.

I have a list of all reciprocal / Pythagorean / quotient identities, its just when I start to play around with it I don't get anywhere.
 
What the heck....this book is wrong then... I don't think it would have a student prove this this /wasn't/ an identity....


Thanks skeeter!
 
jwpaine said:
What are good tricks to proving one side is equal to the other? I start converting either side to one of its identities, and start to get no where in matching to the other side.
Click to expand...

There are no "tricks."

You have several options.

1) work with one side until it looks like the other side

2) work with one side for a while. Then, work with the other side....if you can make it look like what you got after working with the first side, you're done.

Don't be afraid to try things. As another tutor here said earlier, you won't "break it." If you try something that doesn't get you far in the direction you're going, BACK UP and try something else.

Your best bet is to memorize the basic identities (most of which are obtained from the Pythagorean Identity sin<SUP>2</SUP>@+ cos<SUP>2</SUP>@ = 1)
 
maybe it was ...

\(\displaystyle \L \frac{1}{\sec{x}-\tan{x}} = \sec{x} + \tan{x}\) ?
 
skeeter said:
prove that 1/(sec(x)-tan(x)) = csc(x)-sin(x)
Click to expand...

not an identity ...

for x = pi/4, left side is sqrt(2) + 1, right side is sqrt(2)/2
Click to expand...

Why (when my calculator is in radian mode) when I plugin

\(\displaystyle \L \frac{1}{cos^{-1}(pi/4)\,-\,tan(pi/4)}\) = -3.00713

How did you get the left side to equal sqrt(2) + 1 ???
 
jwpaine said:
skeeter said:
prove that 1/(sec(x)-tan(x)) = csc(x)-sin(x)
Click to expand...

not an identity ...

for x = pi/4, left side is sqrt(2) + 1, right side is sqrt(2)/2
Click to expand...

Why (when my calculator is in radian mode) when I plugin

\(\displaystyle \L \frac{1}{cos^{-1}(pi/4)\,-\,tan(pi/4)}\) = -3.00713

How did you get the left side to equal sqrt(2) + 1 ???
Click to expand...

Um......

It is true that sec x = 1/cos x

It is NOT TRUE that sec x = cos<SUP>-1</SUP> x

You may have to get some clarification from your teacher here.
 
I new secant was the inverse of cosine, = 1/cosine but that the cos^-1 was the inverse cosine function which represented this identity?
 
On a calculator \(\displaystyle \cos ^{ - 1} \left( x \right)\) means arccosine of x.
On the other hand, \(\displaystyle \left( {\cos \left( x \right)} \right)^{ - 1} = \sec (x)\)
That is also true for most computer algebra systems.
 
that makes perfect sense pka... thanks.

So the arc functions are to find the angle of a number...

and cos(x)^-1 = 1/cos(x) = sec(x)

I was stupidly confusing them.

:oops:



John.
 
May I just add one point: this is a true failing of mathematics education.
We have done an exceedingly bad job with function notation!
 
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