Melody

Apology accepted.  Welcome to Web2.0 calc forum.   

No you didn't tell me

You said:

tan2x-1 (numerator) / sec2x (denominator) = tanx - cotx (numerator) / tanx + cotx (denominator) 

Maybe this means

A       tan2x- (1  / sec2x) = tanx -   [cotx / (tanx + cotx )]

or

B       (tan2x-1 )  / sec2x  = (tanx - cotx ) / (tanx + cotx)

Is it A or is it B ?

It is important that you learn to go between fractions and brackets.

If you are still confused then take a pic but I would like you to tell me if it is A or B anyway.  :)

I've already tried to help you, respond on the original queston.  

Please post your repaired question here. 

Don't just ignore me and repeat the question elsewhere, that is rude.

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This is what you said in your new question:

I have to vertify the identity (there are no brackets in this problem) 

tan2x-1 (numerator) / sec2x (denominator) = tanx - cotx (numerator) / tanx + cotx (denominator) 

I still do not know what your question is, 

for example, is the numerator  tan2x-1  or     is it   -1   with the 2x seperate?   

That is what brackets are for.  Brackets would make the meaning clear.

If you have copied this from something written with fraction lines then the fraction line is used instead of brackets, if you take away the fraction line you must introduce brackets.

Should there be any brackets here?

Please make it clear what the numerator and denominators are.

Nice answer Textot    

I answered a similar question, not the one asked. (Although the question has been changed to another similar question)

I answered like this because I did not believe the answer to the exact question would be the best way to help this student to learn.

He/she obviously lacks a primary school, pre-algebra, concept.  So I did not want to use algebra to answer.

Thanks Chris.  

You are welcome.

I should have seen the answer straight away but I didn't.   

Maybe I will next time....    

anyway, I am glad I could help.

y=k(x+14)(x-10)+10

Sub in (20,7) to find k

From which end?

18C5  *  (3x)^5  *  z^13

or

18C5  *  (3x)^13  *  z^5