Mary Reyes

2022-01-01

How am I supposed to expand ${\mathrm{sin}}^{2}A+{\mathrm{sin}}^{4}A=1$ into $1+{\mathrm{sin}}^{2}A={\mathrm{tan}}^{2}A$?

Paul Mitchell

${\mathrm{sin}}^{2}A+{\mathrm{sin}}^{4}A=1$
${\mathrm{sin}}^{4}A=1-{\mathrm{sin}}^{2}A$
${\mathrm{sin}}^{2}A{\mathrm{sin}}^{2}A={\mathrm{cos}}^{2}A$
${\mathrm{sin}}^{2}A\left(1-{\mathrm{cos}}^{2}A\right)={\mathrm{cos}}^{2}A$
${\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}A={\mathrm{cos}}^{2}A$
${\mathrm{sin}}^{2}A={\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}A$
${\mathrm{sin}}^{2}A={\mathrm{cos}}^{2}A\left(1+{\mathrm{sin}}^{2}A\right)$
$1+{\mathrm{sin}}^{2}A=\frac{{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}$
$1+{\mathrm{sin}}^{2}A={\mathrm{tan}}^{2}A$

soanooooo40

Hint
If ${\mathrm{sin}}^{2}A+{\mathrm{sin}}^{4}A=1⇒{\mathrm{sin}}^{4}A={\mathrm{cos}}^{2}A$
Can you proceed from here?
${\mathrm{sin}}^{2}A\left(1-{\mathrm{cos}}^{2}A\right)={\mathrm{cos}}^{2}A$
${\mathrm{sin}}^{2}A={\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}A$
Divide by ${\mathrm{cos}}^{2}A$

Vasquez

$\begin{array}{}{\mathrm{sin}}^{2}A=1-{\mathrm{sin}}^{4}A\\ {\mathrm{sin}}^{2}A=\left(1-{\mathrm{sin}}^{2}A\right)\left(1+{\mathrm{sin}}^{2}A\right)\\ {\mathrm{sin}}^{2}A=\left({\mathrm{cos}}^{2}A\right)\left(1+{\mathrm{sin}}^{2}A\right)\\ \therefore {\mathrm{tan}}^{2}A=1+{\mathrm{sin}}^{2}A\end{array}$