Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity.
(x=theta BTW)
(secx/sinx)*(cotx/cscx)
= (secx/cscx)(cotx/sinx)
= (sinx/cosx)*cotx*(1/sinx)
The last steps should be obvious
To verify whether (secx/sinx)*(cotx/cscx) = cscx is an identity, we need to simplify both sides of the equation and show that they are equal.
Let's start by simplifying each side of the equation separately:
Left-hand side (LHS):
(secx/sinx)*(cotx/cscx)
To simplify this expression, we can simplify each fraction individually, and then multiply the resulting terms together. Recall the following trigonometric identities:
secx = 1/cosx
cotx = 1/tanx = cosx/sinx
cscx = 1/sinx
Using these identities, we can rewrite the LHS expression as:
(secx/sinx)*(cotx/cscx) = (1/cosx)*(cosx/sinx)*(1/sinx)
Now, we can cancel out the common terms:
(1/cosx)*(cosx/sinx)*(1/sinx) = (1/sinx) * (1/sinx)
Multiplying the two terms together:
(1/sinx) * (1/sinx) = 1/(sinx * sinx) = 1/sin^2(x) = csc^2(x)
Right-hand side (RHS):
cscx
To simplify the RHS, we can rewrite cscx as 1/sinx:
cscx = 1/sinx
Now that we have simplified both the LHS and RHS, we can compare them:
LHS = csc^2(x)
RHS = cscx
Since csc^2(x) and cscx are equivalent, we can conclude that (secx/sinx)*(cotx/cscx) = cscx is indeed an identity.
By simplifying both sides, we showed that the LHS and RHS are equal to each other, which validates the given identity.