Question #8de97

2 Answers
Dec 10, 2016

That is not an identity.

Explanation:

Recall that

#cot^2x+1 = csc^2x#.

So, we can write

#(1-csc^2x)/csc^2x = (1-(cot^2x+1))/csc^2x#

# = cot^2x/csc^2x#

Recall also that #cotx = cosx/sinx# and #cscx = 1/sinx#.

This allows us to continue

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

# = cos^2x/sin^2x * sin^2x/1#

# = cos^2x#

Which is not identically #cosx#.

(#cos^2x = cosx# only when #cosx = 1# or #0#)

Dec 10, 2016

No. It is equal to #sin^2x - 1#.

Explanation:

If we have #(1 - x)/x#, we can write it as #1/x - x/x#.

The same way, #(1 - csc^2 x)/csc^2# can be written as #1/csc^2x - (csc^2x)/(csc^2x)#.

This is equal to #sin^2x - 1#.