Susan Nall

2022-01-14

For $\alpha \in \left({0}^{\circ };{90}^{\circ }\right)$ simplify ${\mathrm{sin}}^{2}\alpha +{\mathrm{tan}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha \cdot {\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{4}\alpha$
My try: ${\mathrm{sin}}^{2}\alpha +{\mathrm{tan}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha \cdot {\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{4}\alpha ={\mathrm{sin}}^{2}\alpha +\frac{{\mathrm{sin}}^{2}\alpha }{{\mathrm{cos}}^{2}\alpha }+{\mathrm{sin}}^{2}\alpha \cdot {\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{4}\alpha$
$=\frac{{\mathrm{sin}}^{2}\alpha \cdot {\mathrm{cos}}^{2}\alpha +{\mathrm{sin}}^{2}+{\mathrm{sin}}^{2}\alpha \cdot {\mathrm{cos}}^{4}\alpha +{\mathrm{cos}}^{6}\alpha }{{\mathrm{cos}}^{2}\alpha }$

poleglit3

Express everything in terms of $\mathrm{cos}\alpha =c$
$\left(1-{c}^{2}\right)+\frac{\left(1-{c}^{2}\right)}{{c}^{2}}+\left(1-{c}^{2}\right){c}^{2}+{c}^{4}$
$=\frac{{c}^{2}-{c}^{4}+1-{c}^{2}+{c}^{4}-{c}^{6}+{c}^{6}}{{c}^{2}}=\frac{1}{{c}^{2}}$

RizerMix

Youre

user_27qwe

Instead of dividing by ${\mathrm{cos}}^{2}\alpha$, you could do the following, ${\mathrm{sin}}^{2}\alpha +{\mathrm{tan}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha {\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{4}\alpha$ $={\mathrm{sin}}^{2}\alpha +{\mathrm{tan}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha \left({\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha \right)$ $={\mathrm{sin}}^{2}\alpha +{\mathrm{tan}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha$ $=1+{\mathrm{tan}}^{2}\alpha$ $={\mathrm{sec}}^{2}\alpha$