(Sinθ - 1)/(1-sin2θ) = (cscθ)/(-cscθ-1)
or
(1-sinθ)/(1+sinθ) = (tanθ-secθ)^2
1) Please write what you mean. You have written $\displaystyle \sin(2\theta)$. I am guessing this is not your intent.
2) Are you acquainted with the Pythagorean Identity? $\displaystyle \cos^{2}(x) + \sin^{2}(x) = 1$?
3) $\displaystyle \csc(\theta) = 1/\sin(\theta)$
T=[sin(¤)-1]/[1-sin(¤)^2]=[sin(¤)-1]/[(1-sin(¤))(1+sin(¤)] as [a^2-b^2=(a+b)(a-b)]
=-1/1+sin(¤) now dividing both numerator and denominator by sin(¤) we get
T=-csc(¤)/1+csc(¤)=csc(¤)/-1-csc(¤). Hence proved.
2))A=(1-sin(¤))/(1+sin(¤)) multiplying both numerator and denominator by 1-cos(¤) we get A=(1-sin(¤))^2/cos(¤)^2 or [(1-sin(¤))/cos(¤)]^2 which is=[sec(¤)-tan(¤)]^2 which is= [tan(¤)-sec(¤)]^2 hence proved.
Note (a-b)^2=(b-a)^2