(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)
[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x]
[(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)
(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)
[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)
(1 - cos x)/sin x =? sin x/(1 + cos x)
(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)]
(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)
(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)]
(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x
(1 - cos x)/sin x = (1 - cos x)/sin x True
Cannot prove that 2 divided by 10 equals 2 because it is not true.
What divided by 7 equals 8? In other words, you have an unknown number (X), and then if you divide that X by 7 you get 8. Then what is that X? The equation to calculate what divided by 7 equals 8 is as follows: X/7 = 8 Where X is the answer. When we solve the equation by multiplying each side by 7, you get get: X = 56 Therefore, the answer to what divided by 7 equals 8 is 56.
Because there is no way to define the divisors, the equations cannot be evaluated.
You can't it equals 2. You can't it equals 2.
No you can not prove that 9 +10 = 21.
It is not possible to prove it because it is not true!
No, but there is a way to prove that zero equals one.
Using faulty logic.
You cannot prove it since it is axiomatic. You can get consistent theories (matrix algebra, for example) where ab is not ba.
It is extremely difficult to prove things which are not true.
Using a calculator
4,4,4,4 = 20
This would be a real bear to prove, mainly because it's not true.
that cannot be proved because it is not necessarily true.
It isn't equal, and any proof that they are equal is flawed.
It is REM(4n/6)=4. Prove. Correction.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. 6 is an even number.
There is not much to prove there; opposite numbers, by which I take you mean "additive inverse", are defined so that their sum equals zero.
No, because technically, it is not true.