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.