A)
B)
C)
This is unconventional I'm sure but in regars to a):
(cscx-cotx)^2=csc^2x-2cscxcotx+cot^2x=
(1/sin^2x)-2(cosx/sin^2x)+(cos^2x/sin^2x)=
(1-2cosx+cos^2x)/sin^2x=(1-cosx)/(1+cosx)
multiplying both sides by 1+cosx yields
(1-cosx-cos^2x+cos^3x)/(1-cos^2x)=1-cosx
(remember that sin^2x=1-cos^2x)
Now Multiply both sides by 1-cos^2x
1-cosx-cos^2x+cos^3x=1-cosx-cos^2x+cos^3x
I know that this is probably a strange way to prove this identity, but I learned everything I know about math from a prison library.
Hello, VonNemo19!
The accepted rule is that we never play with both side of the "equation".This is unconventional I'm sure, but in regards to a):
. .
. .
multiplying both sides by . . . . no
It is not an equation because it has not been established that the two sides are equal.
In fact, that is what we are asked to prove. .**
You could continue like this: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Imagine there is philosophy test queston: Prove that there is a God.
And I answer: There is a God, because in the Bible, it says:
. . . . . . . . . . "In the Beginning, God created Heaven and Earth."
. . . . . . . . . . And God would never lie.