Question

(1+sint)/(cost)-(cost)/(1-sint)

Answer 1

$\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{\cos(t)}{1-\sin(t)}$

Multiply through the second term by $1+\sin(t)$ :

$\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{\cos(t)}{1-\sin(t)}\cdot\dfrac{1+\sin(t)}{1+\sin(t)}$

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

Recall that $\cos^2(x)+\sin^2(x)=1$ for all x :

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

If $\cos(t)\neq0$, we can cancel a factor of it in the second fraction.

$\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{1+\sin(t)}{\cos(t)}$

and this simplifies to

$\dfrac{(1+\sin(t))-(1+\sin(t))}{\cos(t)} = \boxed{0}$