2 Trigonometric Identities Questions..can you solve them?

Question

Heres what you work with:
sin^2x+cos^2x=1
1+tan^2x=sec^2x
1+cot^2x=csc^2x

The Questions:
Question 1:

(sinx+cosx÷sinx ) - (cosx-sinx÷cosx)

Question 2:

1+2sinx+sin^2x ÷ cos^2x

Alright Question 1 again:
((sinx+cosx)÷(sinx)) - ((cosx-sinx)÷(cosx))

Answer 1

You should be more explicit with your bracketing.

For Q1 I don't know if you mean
a) sin x + (cos x / sinx) - (cos x - (sin x / cos x))
or
b) (sin x + cos x) / sin x - (cos x - sin x) / cos x

a) = sin x + (cos^2x / (sin x cos x)) - cos x + (sin^2 x / (sin x cos x))
= sin x - cos x + 1/(sin x cos x)

b) = (sin x cos x + cos^2 x) / (sin x cos x) - (cos x sin x - sin^2 x) / (sin x cos x)
= (sin x cos x + cos^2 x - cos x sin x + sin^2 x) / (sin x cos x)
= 1 / (sin x cos x)

2. I assume this is supposed to be
(1 + 2sin x + sin^2 x) / cos^2 x.
= (1 + sin x)^2 / (1 - sin^2 x)
= (1 + sin x)^2 / [(1 - sin x) (1 + sin x)]
= (1 + sin x) / (1 - sin x)

Answer 2

use a graphing calculator. i would help, but im scared ill giv u wrong answer. i flunked trig. gud luck its a monster!!!!!!