cos x cot = csc - sin
What are we doing?
Solving ?
or
proving it is an identity?
BTW, you have to put the x for all the trig functions
that is,
cosx cotx = cscx - sinx
Let's see if it is an identity...
LS = cosx (cosx/sinx
= cos^2 x/sinx
RS = 1/sinx - sinx
= (1 - sin^2 x)/sin
= cos^2 x/sinx
= LS
ahh, it is an identity
Cos (cos/sin)= cos^2/sin = (1 - sin^2x)/sin
1/sin sin^2/sin = csc x sin x
To simplify the expression cos x cot = csc - sin, we need to use the trigonometric identities and algebraic manipulation.
First, let's rewrite the given equation in terms of sine and cosine:
cos x cot = csc - sin
Since cot(x) is the reciprocal of tan(x), we can write it as 1/tan(x):
cos x (1/tan x) = csc - sin
Next, let's express csc(x) in terms of sine:
csc x = 1/sin x
Now we have:
cos x (1/tan x) = 1/sin x - sin
Using the fact that tan(x) = sin(x)/cos(x), we can simplify further:
cos x (cos x / sin x) = 1/sin x - sin
cos^2 x / sin x = 1/sin x - sin
To eliminate the fractions, we can multiply both sides of the equation by sin x:
cos^2 x = 1 - sin^2 x
Applying the Pythagorean identity for cosine and sine, which states that cos^2 x + sin^2 x = 1, we can substitute:
1 - sin^2 x = 1 - sin^2 x
This equation is true for all values of x.
Therefore, the original equation cos x cot = csc - sin holds true for any value of x.