Expression with parametric formulae

Problem

Using the parametric formulae, express the expression $\frac{1+\sin\alpha}{\cos\alpha}$ in terms of $t=\tan(\alpha/2)$ and simplify it.

Solution

We substitute $\sin\alpha = \dfrac{2t}{1+t^2}$ and $\cos\alpha = \dfrac{1-t^2}{1+t^2}$. In the numerator: $1+\sin\alpha = \frac{1+t^2+2t}{1+t^2} = \frac{(1+t)^2}{1+t^2}.$ Dividing by the cosine: $\frac{1+\sin\alpha}{\cos\alpha} = \frac{(1+t)^2}{1+t^2}\cdot\frac{1+t^2}{1-t^2} = \frac{(1+t)^2}{1-t^2} = \frac{(1+t)^2}{(1-t)(1+t)} = \boxed{\dfrac{1+t}{1-t}}.$

Links

Topics: Prostaferesi werner
Concepts: Formule parametriche · Tangente dell angolo meta
Functions: Coseno · Seno
Skills: Semplificare · Usare formule
Exercise type: Calcolo espressioni