prove:

cot^2(x)=cos(x)/(sinx)(tanx)

tan ( x ) = sin ( x ) / cos ( x )

sin ( x ) * tan ( x ) ) sin ( x ) * sin ( x ) / cos ( x ) = sin ^ 2 ( x ) / cos ( x )

cos( x )/ [ sin ( x ) * tan ( x ) ]=

cos ( x ) / [ sin ^ 2 ( x ) / cos ( x ) ] =

[ cos ( x ) / 1 ] / [ sin ^ 2 ( x ) / cos ( x ) ]=

cos ( x ) * cos ( x ) / sin ^ 2 ( x ) * 1 =

cos ^ 2 ( x ) / sin ^ 2 ( x ) = ctg ^ 2 ( x =

sin ( x ) * tan ( x ) = sin ( x ) * sin ( x ) / cos ( x ) = sin ^ 2 ( x ) / cos ( x )

To prove the given identity, we need to find a way to simplify the left-hand side (LHS) and the right-hand side (RHS) of the equation separately, and then show that they are equal.

We can start by simplifying the LHS, which is cot^2(x):

Recall that cot(x) = cos(x) / sin(x). Substituting this value into the LHS, we have:

cot^2(x) = (cos(x) / sin(x))^2

To square a fraction, we square the numerator and the denominator separately:

cot^2(x) = (cos^2(x)) / (sin^2(x))

Now let's simplify the RHS, which is cos(x) / (sin(x) * tan(x)):

Recall that tan(x) = sin(x) / cos(x). Substituting this value into the denominator, we get:

cos(x) / (sin(x) * (sin(x) / cos(x)))

Now, simplifying further:

cos(x) / (sin(x) * (sin(x) / cos(x))) = cos(x) / (sin^2(x) / cos(x))

To divide by a fraction, we can invert and multiply:

cos(x) / (sin^2(x) / cos(x)) = (cos(x) * cos(x)) / sin^2(x)

We can simplify further by using the Pythagorean identity: sin^2(x) + cos^2(x) = 1. Rearranging this equation, we have:

cos^2(x) = 1 - sin^2(x)

Substituting this value into the RHS, we get:

(cos(x) * cos(x)) / sin^2(x) = (1 - sin^2(x)) / sin^2(x)

Now, let's compare the simplified expressions for the LHS and RHS:

LHS: cot^2(x) = (cos^2(x)) / (sin^2(x))
RHS: (1 - sin^2(x)) / sin^2(x)

Since the LHS and RHS have been simplified to the same expression, we have proven that cot^2(x) is equal to cos(x) / (sin(x) * tan(x)).