Alex Cervantes

2022-01-24

Prove that $\mathrm{cos}e{c}^{2}A\cdot {\mathrm{sec}}^{2}A={\mathrm{tan}}^{2}A+{\mathrm{cot}}^{2}A+2$

Tapanuiwp

Expert

I assume your second ${\mathrm{csc}}^{2}A$ was meant to be ${\mathrm{sec}}^{2}A$, viz.
${\mathrm{tan}}^{2}A+{\mathrm{cot}}^{2}A+2={\left(\mathrm{tan}A+\mathrm{cot}A\right)}^{2}={\left(\frac{{\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A}{\mathrm{sin}A\mathrm{cos}A}\right)}^{2}$
$=\frac{1}{{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}A}={\mathrm{csc}}^{2}A{\mathrm{sec}}^{2}A$

pripravyf

Expert

Let . Then $\mathrm{tan}A=\frac{s}{c}$ and $\mathrm{cot}A=\frac{c}{s}$. Therefore
$RHS={\mathrm{tan}}^{2}A+{\mathrm{cot}}^{2}A+2={\left(\frac{s}{c}\right)}^{2}+{\left(\frac{c}{s}\right)}^{2}+2$
This gives
$RHS=\frac{{s}^{4}+{c}^{4}+2{s}^{2}{c}^{2}}{{s}^{2}{c}^{2}}=\frac{{\left({s}^{2}+{c}^{2}\right)}^{2}}{{s}^{2}{c}^{2}}$
But ${s}^{2}+{c}^{2}={\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A=1$. Therefore
$RHS=\frac{1}{{s}^{2}{c}^{2}}={\left(\frac{1}{s}\right)}^{2}{\left(\frac{1}{c}\right)}^{2}$
Since $\mathrm{cos}ecA=\frac{1}{s}$ and $\mathrm{sec}A=\frac{1}{c}$, the claim follows (presumably you wanted to prove $\mathrm{cos}e{c}^{2}A{\mathrm{sec}}^{2}A={\mathrm{tan}}^{2}A+{\mathrm{cot}}^{2}A+2\right)$

Do you have a similar question?