[tex]\dfrac{\cos^4t-\sin^4t}{\sin^2t}=\dfrac{(\boxed{\cos^2t}+\sin^2t)(\cos^2t-\sin^2t)}{\sin^2t}[/tex]
which follows from the difference-of-squares identity, [tex]a^2-b^2=(a+b)(a-b)[/tex], with [tex]a=\cos^2t[/tex] and [tex]b=\sin^2t[/tex].
[tex]=\dfrac{(\boxed{1})(\cos^2t-\sin^2t)}{\sin^2t}[/tex]
which is due to the Pythagorean identity.
[tex]=\dfrac{\boxed{\cos^2t}}{\sin^2t}-\dfrac{\sin^2t}{\sin^2t}[/tex]
by the distributive property; [tex]1(\cos^2t-\sin^2t)=\cos^2t-\sin^2t[/tex].
[tex]=\cot^2t-1[/tex]
by definition of cotangent, [tex]\cot t=\frac{\cos t}{\sin t}[/tex].
[tex]\dfrac{\cos^2\theta}{1-\sin\theta}=\dfrac{1-\boxed{\sin^2\theta}}{1-\sin\theta}[/tex]
due to the Pythagorean identity.
[tex]=\dfrac{(1-\boxed{\sin\theta})(1+\sin\theta)}{1-\sin\theta}[/tex]
by factorization of the numerator as a difference of squares.
[tex]=1+\sin\theta[/tex]
by cancellation of [tex]1-\sin\theta[/tex] (provided [tex]1-\sin\theta\neq0[/tex]).
