Prove this!

[prog_tom]

(tanx / (1 - cot x)) + (cotx/(1-tanx)) = tanx + cotx + 1

So far I have gotten these:

(tanx - tanx^2 + cotx - cotx^2)/(1-tanx+1-cotx)

divide top and bottom i get:

tan(2 - 2tan) + cot(2 - 2cot)

again, divide by 2

tan + cot - tan^2 - cot^2

[prog_tom]

btw, anyone know what sine * cos is?

[sql_lall]

Let tan(x) = A, cot(x) = 1/A

RTP: A/(1-1/A) + (1/A)/(1-A) = A + 1/A + 1

(2- A - 1/A)(A + 1/A + 1)
= (A + 1/A - A² - 1/A²)

=> LHS = (A + 1/A - A² - 1/A²)/(2 - A - 1/A) = A + 1/A + 1 = RHS

N.B. to get LHS = (A + 1/A - A² - 1/A²)/(2 - A - 1/A), just put over common denominator (1-1/A)(1-A) = (2-A-1/A)

[TheAlchemist]

hey sql_lall
could you please explain how you got
( A + 1/A - A^2 - 1/A^2)/ (2- A - 1/A)
is equal to A + 1/A +1 ?

[sql_lall]

(2- A - 1/A)(A + 1/A + 1)
=2A + 2/A + 2 - A^2 - 1 - A - 1 - 1/A^2 - 1/A
=(A + 1/A - A^2 - 1/A^2)

=>(2 - A - 1/A)(A + 1/A + 1) = (A + 1/A - A^2 - 1/A^2)

=> (A + 1/A - A^2 - 1/A^2)/(2 - A - 1/A) = (A + 1/A + 1)
-dividing both sides by (2 - A - 1/A)

This is a good case of "knowing" the answer is right (otherwise, why are you trying to proove it )
Then, upon getting a trick fraction, (A + 1/A - A^2 - 1/A^2)/(2 - A - 1/A), you know it *should* = (A + 1/A + 1)

=> to proove this, show (A + 1/A + 1)*(2 - A - 1/A) = (A + 1/A - A^2 - 1/A^2) Q.E.D

[bugzpodder]

prog_tom, no but i know what cos+sine is! it is cossine!!

[siyan]

Originally posted by bugzpodder
prog_tom, no but i know what cos+sine is! it is cossine!!

o ho ho ho ho

[sql_lall]

well, i'm not sure what it has to do with anything, but:

either:
1) sin(X) = sqrt(1-cos²(X))
=> sin(X) + cos(X) = sqrt(1-cos²(X)) + cos(X), which isn't really all that good, but anyway.

2) tan(X) = sin(X)/cos(X)
=> sin(X) = tan(X)cos(X)
=> cos(X)+sin(X) = cos(X)(tan(X)+1)
which looks a bit nicer