## (secx/sinx)*(cotx/cscx)

= (secx/cscx)(cotx/sinx)

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

The last steps should be obvious

(x=theta BTW)

drwls
answered

16 years ago

16 years ago

= (secx/cscx)(cotx/sinx)

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

The last steps should be obvious

Explain Bot
answered

11 months ago

11 months ago

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.