Naar aanleiding van Bewijzen van gelijkheden:
Opdracht 1
\(
\eqalign{
& \cos \alpha \cdot \left( {\tan \alpha + 2} \right) \cdot \left( {2\tan \alpha + 1} \right) = \frac{2}
{{\cos \alpha }} + 5\sin \alpha \cr
& \left( {\tan \alpha + 2} \right) \cdot \left( {2\tan \alpha + 1} \right) = \frac{2}
{{\cos ^2 \alpha }} + 5\tan \alpha \cr
& 2\tan ^2 \alpha + 5\tan \alpha + 2 = \frac{2}
{{\cos ^2 \alpha }} + 5\tan \alpha \cr
& 2\tan ^2 \alpha + 2 = \frac{2}
{{\cos ^2 \alpha }} \cr
& \frac{{\sin ^2 \alpha }}
{{\cos ^2 \alpha }} + 1 = \frac{1}
{{\cos ^2 \alpha }} \cr
& \frac{{\sin ^2 \alpha }}
{{\cos ^2 \alpha }} + \frac{{\cos ^2 \alpha }}
{{\cos ^2 \alpha }} = \frac{1}
{{\cos ^2 \alpha }} \cr
& \frac{{\sin ^2 \alpha + \cos ^2 \alpha }}
{{\cos ^2 \alpha }} = \frac{1}
{{\cos ^2 \alpha }} \cr
& \frac{1}
{{\cos ^2 \alpha }} = \frac{1}
{{\cos ^2 \alpha }} \cr}
\)
Opdracht 2
\(
\eqalign{
& \frac{1}
{{1 - \sin \alpha }} + \frac{1}
{{1 + \sin \alpha }} = \frac{2}
{{\cos ^2 \alpha }} \cr
& \frac{1}
{{1 - \sin \alpha }} \cdot \frac{{1 + \sin \alpha }}
{{1 + \sin \alpha }} + \frac{1}
{{1 + \sin \alpha }} \cdot \frac{{1 - \sin \alpha }}
{{1 - \sin \alpha }} = \frac{2}
{{\cos ^2 \alpha }} \cr
& \frac{{1 + \sin \alpha + 1 - \sin \alpha }}
{{1 - \sin ^2 \alpha }} = \frac{2}
{{\cos ^2 \alpha }} \cr
& \frac{2}
{{1 - \sin ^2 \alpha }} = \frac{2}
{{\cos ^2 \alpha }} \cr
& \frac{2}
{{\cos ^2 \alpha }} = \frac{2}
{{\cos ^2 \alpha }} \cr}
\)
Bij wiskunde komt altijd alles wel weer ergens terug...:-)
