# How to prove the following: tan^2x+1+tanx secx = 1+sin x/ cos^2x

Cem Hayes 2021-03-02 Answered
How to prove the following:
${\mathrm{tan}}^{2}x+1+\mathrm{tan}x\mathrm{sec}x=1+\frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}$
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hosentak
$\mathrm{tan}x\cdot \mathrm{sec}x=\frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{sin}x\cdot {\mathrm{sec}}^{2}x.{\mathrm{tan}}^{2}x+1={\mathrm{sec}}^{2}x.$
So we have ${\mathrm{sec}}^{2}x+{\mathrm{sec}}^{2}x\mathrm{sin}x={\mathrm{sec}}^{2}x\left(1+\mathrm{sin}x\right)=\frac{1+\mathrm{sin}x}{{\mathrm{cos}}^{2}x}.$
(To prove ${\mathrm{sec}}^{2}x={\mathrm{tan}}^{2}x+1$, start with ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$, then divide through by ${\mathrm{cos}}^{2}x:{\mathrm{tan}}^{2}x+1={\mathrm{sec}}^{2}x$ because secx is $\frac{1}{\mathrm{cos}x}$.)