Please help me in verifying Trig Identities:

Question

1.( 2-cos^2x)/sinx = cscx+sinx

2. (1-cscx)/cscx=sinx-1/sinx+1

3. csc^2x-cos^2x-sin^2x=cot^2x?

Answer 1

1) (2 - cos²x)/sinx = cscx + sinx

Left hand side identity = (2 - cos²x)/sinx

[Recall the Pythagorean identities sin²x + cos²x = 1]

= [2 - (1 - sin²x)]/sinx

= (2 - 1 + sin²x)/sinx

= (1 + sin²x)/sinx

= 1/sinx + sin²x/sinx

= cscx + sinx

= Right hand side identity.

Answer 2

3) csc²x - cos²x - sin²x = cot² x

Left hand side identity = csc²x - cos²x - sin²x

{Recall the Pythagorean identities sin²x + cos²x = 1 and reciprocal identity of sinx is cscx}

= (1/sin²x) - ( 1 - sin²x ) - sin²x

= (1/sin²x) - 1 + sin²x  - sin²x

= (1/sin²x) - 1

= (1 - sin²x)/sin²x

= cos²x/sin²x

= cot²x

= Right hand side identity.

Answer 3

2) Assume the identity ( 1 - cscx)/(1 + cscx) = (sinx - 1)/(sinx + 1)

Left hand side identity = (1 - cscx)/(1 + cscx)

{Reciprocal identity of cscx is sinx }

= [ 1 - (1/sinx)]/[1 + (1/sinx)]

= [(sinx - 1)/sinx]/[(sinx + 1)/sinx]

= (sinx - 1)/(sinx + 1)

= Right hand side identity.