Kassandra Ross

2022-06-28

Proving $1+{\mathrm{tan}}^{2}{40}^{\circ }={\mathrm{sec}}^{2}{40}^{\circ }$

jarakapak7

Expert

$1+{\mathrm{tan}}^{2}x=1+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}=:{\mathrm{sec}}^{2}x.$
Yours is just the particular case x=40.

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