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.