Question

cos/(sec-1)-cos/tan^2=cot^2

1/(1-sin)=sec^2+sec•tan

If someone could explain these to me I'd be forever grateful!

Answer 1

is the den'r of that first term really sec(x)-1? or is it \(sec^{-1}(x)\)?

And the trig functions without a variable really drives me crazy, lol. A function with no input is mathematically gobbledygook! ;)

Answer 2

Ok. I'll rewrite it. cos(a)/(sec(a)-1)-cos(a)/tan(a)^2=cot(a)^2

1/(1-sin(a))=sec(a)^2+sec(a)•tan(a)

Answer 3

And to answer your question, yes it is actually sec(x)-1.

Answer 4

OK, you need to show the steps to prove that the expression on the left is = the expression on the right side.

Here is the "trick" you need with this one (you can sometimes spot the ones that need an algebraic trick... here it was because the we need squared terms, and we have that difference in that first den'r that does not involve a square).

Multiply that whole first term by: \(\Large \dfrac {\sec x+1}{\sec x+1}\)

When you do that and then some simplification, and use the pythagorean identity:

\(\Large \tan^2 x+1=\sec^2 x\) in the process.... it will all fall into place. :)

Answer 5

Please explain what sec(x)+1(cos(x)) would end up being.

Answer 6

On the 2nd one, start on the RHS. If you rewrite the first term as 1/cos^2(x), and simplify the 2nd term by converting it to sines and cosines, you'll have an LCD and can combine the RHS over that den'r. Then again, you'll do a multiplication "trick" similar to in the first one (multiply by (1-sinx)/(1-sinx)). Then you'll see some things that can be cancelled and you'll have what's on the left side.

Answer 7

\(\Large (\dfrac {\cos x}{\sec x-1})(\dfrac {\sec x+1}{\sec x+1})\)
\(\Large =\dfrac {\cos x(\sec x+1)}{\sec^2 x-1}\)
\(\Large =\dfrac {1+\cos x}{\sec^2 x-1}\)

Now you can make a substitution in the den'r, using:
\(\Large \tan^2 x+1=\sec^2 x\)

After that substituion, you'll be able to add the other term, because you will have a common den'r.

Then watch what happens in the numerator.

Answer 8

Oooo! because \[\cos x \times \sec x =1\]

Answer 9

That's right. :) handy, huh? lol

Answer 10

I'm going to have to go - I don't want to desert you, but I think you're on the right track now. you have:

\(\Large \dfrac {1+\cos x}{\sec^2 x-1}-\dfrac {\cos x}{\tan^2 x}\)

Now use the pyth ID to sub \(\tan^2 x=\sec^2 x-1\) in that first den'r.

Add the num'rs over the den'r.

and you should be pretty much there. :)

The other one uses similar methods... just follow my hint above. :)

Answer 11

Yes:) now for the second one I got up to \[\frac{ 1 }{ 1-\sin x }=\frac{ \sin x +1}{ \cos ^{2}x }\left(\begin{matrix}\sin x-1 \\ \sin x-1\end{matrix}\right)\]
The numerator I got sin^2(x)-1
I don't understand the denominator.