Prove the identity:

1-cos2x/sin2x = 1+tanx/1+cotx

I simplified the RS to tanx.
LS:(1-2cos²x+1)/2sinxcosx
(2+2cos²x)/2sinxcosx

How do I simplify the rest?

change the LS to

(1 - (1 - 2sin²x))/(2sinxcosx)
= 2sin²x/(2sinxcox)
= sinx/cosx
= tanx = RS

thanks! I've been trying to get this one for quite awhile. So many options to choose :(

To simplify the remaining expression, we can start by simplifying the numerator and denominator separately.

Numerator:
The numerator of the expression is 2 + 2cos²x. We can factor out a 2:
2(1 + cos²x).

Note that 1 + cos²x is a well-known trigonometric identity:
1 + cos²x = sin²x.

Therefore, the numerator simplifies to:
2sin²x.

Denominator:
The denominator is 2sinxcosx.

Now, let's substitute these simplified forms back into the original expression:
(2sin²x) / (2sinxcosx).

We can cancel out the common factor of 2 from the numerator and denominator:
(sin²x) / (sinxcosx).

Next, we can simplify further using trigonometric identities:
sinxcosx = 1/2 * 2sinxcosx = 1/2 * 2sinxcosx = 1/2 * sin2x.

Substituting this back into the expression, we get:
(sin²x) / (1/2 * sin2x).

Now, we can simplify by multiplying the numerator by the reciprocal of the denominator:
(sin²x) * (2/sin2x).

The sin2x terms in the numerator and denominator cancel out, leaving:
(sin²x) * (2/1).

Finally, simplifying this expression further gives us the result:
2sin²x.

Therefore, the simplified form of the original expression is 2sin²x, which is the same as tan(x) (as you previously derived). This confirms the given identity.