Question

Can anyone help me with these two problems? Thank you.

Answer 1

$\dfrac{\cos^4t-\sin^4t}{\sin^2t}=\dfrac{(\boxed{\cos^2t}+\sin^2t)(\cos^2t-\sin^2t)}{\sin^2t}$

which follows from the difference-of-squares identity, $a^2-b^2=(a+b)(a-b)$, with $a=\cos^2t$ and $b=\sin^2t$.

$=\dfrac{(\boxed{1})(\cos^2t-\sin^2t)}{\sin^2t}$

which is due to the Pythagorean identity.

$=\dfrac{\boxed{\cos^2t}}{\sin^2t}-\dfrac{\sin^2t}{\sin^2t}$

by the distributive property; $1(\cos^2t-\sin^2t)=\cos^2t-\sin^2t$.

$=\cot^2t-1$

by definition of cotangent, $\cot t=\frac{\cos t}{\sin t}$.

$\dfrac{\cos^2\theta}{1-\sin\theta}=\dfrac{1-\boxed{\sin^2\theta}}{1-\sin\theta}$

due to the Pythagorean identity.

$=\dfrac{(1-\boxed{\sin\theta})(1+\sin\theta)}{1-\sin\theta}$

by factorization of the numerator as a difference of squares.

$=1+\sin\theta$

by cancellation of $1-\sin\theta$ (provided $1-\sin\theta\neq0$).