Trigonometric Identities

can someone help me with my math homework please i dont understand these 2 questions

Prove that each of the following equations is an identity

1) 1+sinΘ / sinΘ = cosΘcotΘ /1-sinΘ

2) secΘ - cosΘ = (cosΘsinΘ / 1-sinΘ ) - tanΘ

1) Working with right side:
cosΘcotΘ /1-sinΘ =
cosΘ(cosΘ/sinΘ) / 1-sinΘ =
[(cosΘ)^2 / sinΘ ] / 1-sinΘ =
[(1 - (sinΘ)^2) / sinΘ] / 1-sinΘ =
(1 - (sinΘ)^2) / [(sinΘ)(1-sinΘ)] =
[(1 + sinΘ)(1-sinΘ)] / [(sinΘ)(1-sinΘ)] =
(1 + sinΘ) / (sinΘ)=
1+sinΘ / sinΘ = 1+sinΘ / sinΘ

2) Working with the right side:
(cosΘsinΘ / 1-sinΘ ) - tanΘ =
[cosΘ(tanΘcosΘ) / 1-sinΘ ] - tanΘ =
[(cosΘ)^2(tanΘ) / 1-sinΘ ] - tanΘ =
[(cosΘ)^2 / 1-sinΘ](tanΘ) - tanΘ =
[(1-sinΘ^2) / 1-sinΘ](tanΘ) - tanΘ =
[(1+sinΘ)(1-sinΘ)/ 1-sinΘ](tanΘ) - tanΘ =
(1+sinΘ)(tanΘ) - tanΘ =
tanΘ (1 +sinΘ - 1) =
tanΘsinΘ =
(sinΘ/cosΘ)(sinΘ) =
(sinΘ)^2/cosΘ =
(1 - (cosΘ)^2) / cosΘ =
1/cosΘ - (cosΘ)^2/cosΘ =
secΘ - cosΘ
EDIT: secΘ - cosΘ = secΘ - cosΘ

hope this helps!!! :)

1)

(1 + sin A) / sin A

multiply top and bottom with (1 - sin A)

(1 - sin^2(A) /sin A (1 - sin A)

= cos^2(A) / sin A (1 - sin A)

= ( cos A)( cosA / sin A) / (1 - sin A)

= (cos A cot A ) /(1 - sin A)

2)

take RHS

{ ( cos A sin A) / (1 - sin A) } - tan A

= (1 + sin A){ ( cos A sin A) / (1 - sin^2(A) } - tan A

= (1 + sin A){ ( cos A sin A) /cos^2(A) } - tan A

= (1 + sin A) sin A) /cos(A) } - tan A

= tan A (1 + sin A) - tan A

= tan A { 1 + sin A - 1 }

= tan A sin A

= sin^2(A) / cos A

= (1 - cos^2(A) / cos A

= (1 / cos A ) - cos A

= sec A - cos A