Verify the identity

cot2 beta/1+csc beta=1-sin beta/sin beta.

did you mean

cot^2 ß/(1+cscß) = (1-sinß)/sinß ?

from the identity
sin^ Ø + cos^2 Ø = 1
if we divide each term by sin^2Ø ......
1 + cot^2 Ø = csc^2 Ø ----> cot^2 Ø = csc^2Ø - 1

LS
= (csc^2 ß - 1)/(1 + csc ß)
= (cscß + 1)(cscß - 1)/(1 + cscß)
= cscß - 1
= 1/sinß - sinß/sinß
= (1 - sinß)/sinß
= RS

To verify the given identity:

cot²(β) / (1 + csc(β)) = (1 - sin(β)) / sin(β)

We will manipulate the left-hand side (LHS) and the right-hand side (RHS) separately to see if they can be simplified to the same expression.

Starting with the LHS:

cot²(β) / (1 + csc(β))

Using the reciprocal identities, we can express cotangent and cosecant in terms of sine and cosine:

= (cos²(β) / sin²(β)) / (1 + (1 / sin(β)))

Simplifying the denominator by finding a common denominator:

= (cos²(β) / sin²(β)) / ((sin(β) + 1) / sin(β))

Next, we invert the denominator and multiply:

= (cos²(β) / sin²(β)) * (sin(β) / (sin(β) + 1))

Simplifying the expression by canceling like terms:

= cos²(β) / (sin(β) + 1)

Now, let's simplify the RHS:

(1 - sin(β)) / sin(β)

Expanding the numerator:

= 1/sin(β) - sin(β)/sin(β)

Simplifying the denominator:

= 1/sin(β) - 1

Using the reciprocal identity for sine:

= csc(β) - 1

Therefore, we have:

= cos²(β) / (sin(β) + 1) = csc(β) - 1

We have shown that the LHS is equal to the RHS, thereby verifying the identity.