My previous question:

Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity.

(secx/sinx)*(cotx/cscx)
= (secx/cscx)(cotx/sinx)
= (sinx/cosx)*cotx*(1/sinx)

"The last steps should be obvious"

Not to me. I can convert (sinx/cosx) to tanx if that's even what I'm suppose to do. Is (sinx/cosx)*cotx*(1/sinx) the answer?

I am not sure he is still here

(sinx/cosx)*cotx*(1/sinx)
(sin/cos)*(cos/sin) * (csc)

1 * (csc) = csc

remember csc = 1/sin

I skipped that step that tanx*cotx = 1

To verify that (secx/sinx)*(cotx/cscx) = cscx is an identity, we need to simplify both sides of the equation and show that they are equal.

Starting with the left-hand side (LHS):

(secx/sinx)*(cotx/cscx)
= (secx/cscx)*(cotx/sinx) (interchanging the factors)
= (1/cosx)*(cotx/sinx) (since secx = 1/cosx and cscx = 1/sinx)
= cotx/(cosx*sinx) (multiplying the fractions)

Now, let's simplify the right-hand side (RHS):

cscx
= 1/sinx (definition of cscx)

Comparing the LHS and RHS, we have:

cotx/(cosx*sinx) = 1/sinx

To further simplify the LHS, we can rewrite cotx as cosx/sinx:

(cotx/(cosx*sinx)) = (cosx/sinx)/(cosx*sinx)
= (cosx/sinx) * (1/(cosx*sinx)) (multiplying the fractions)

Now, we cancel out the cosx in the numerator and denominator:

(cosx/sinx) * (1/(cosx*sinx)) = 1/sinx

Since we have shown that both sides of the equation simplify to 1/sinx, we can conclude that the given statement (secx/sinx)*(cotx/cscx) = cscx is indeed an identity.