guys make me look good...

[ayoub_ibrahim]

Ok been a looong time since I did this sort of maths so can anyone prove
(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA) and post your equations?

my nephew thinks I should be able to do it.

[GNEP]

IIRC you have to express them in terms of e^i*whatever (you should be able to find the equivalencies easily enough on the interweb) and then it should drop out...

[gt40]

I would have certainly been able to help 25-30 yrs ago, but normal life/work have made my mind mush
might have something to do with the cosine law, possibly

[Mehen]

use the identities found at the bottom of the page http://www.mathematicshelpcentral.co...n_Formulas.pdf
if you still cannot get it then I will try it - but I gotta crack open my psych text for my midterm tomorrow

[the main lobe]

Piece of cake

First distribute and change the inverted sine and cosine functions to regular sine and cosines.
left hand side of the equation...

[1/(sinA*cosA)]-[cosA / sinA]-[sinA / cosA]+cosA*sinA

Next get a common denominator in the middle terms and this becomes

[1/{sinA*cosA)]-[(cos^2 A + sin^2 A)/sinA*cosA] +cosA*sinA

Since (cos^2 A + sin^2 A)=1 the left side of the equation becomes just

cosA*sinA

The right side:

get rid of tan and cot by replacing with (sin/cos) and (cos/sin) respectively. Do the same common denominator step as on the left and you end up with

cosA*sinA=cosA*sinA

[DukeP]

What do you know: One actually DOES learn something in Grad School.

Nicely done, the_main_lobe.

~~DukeP~~

[ayoub_ibrahim]

First distribute and change the inverted sine and cosine functions to regular sine and cosines.
left hand side of the equation...

[1/(sinA*cosA)]-[cosA / sinA]-[sinA / cosA]+cosA*sinA

Next get a common denominator in the middle terms and this becomes

[1/{sinA*cosA)]-[(cos^2 A + sin^2 A)/sinA*cosA] +cosA*sinA

Since (cos^2 A + sin^2 A)=1 the left side of the equation becomes just

cosA*sinA

The right side:

get rid of tan and cot by replacing with (sin/cos) and (cos/sin) respectively. Do the same common denominator step as on the left and you end up with

cosA*sinA=cosA*sinA

excellent stuff mate and thanks a bunch!

[Rakido]

hope that helps ...

[GNEP]

Nice one main lobe Clearly I'm a little rusty...

[rylan]

What do you know: One actually DOES learn something in Grad School.

Nicely done, the_main_lobe.

~~DukeP~~

Learn something yes... learn something useful? No

[Lambo-Fan]

LOL @Rakido

Somehow that picture reminds me of my maths classes in school

[KvHagedorn]

What would remind me of my college math classes would be someone from China or Argentina with absolutely zero communication skills silently working problems on the board for the entire hour and expecting us to learn something from it.

[VJ]

Don't forget to include the cases in which the calculation fails (due to division by zero and such).

(they found this extremely important at our university)


Jörg