Alex Cervantes

Answered

2022-01-24

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

Answer & Explanation

Tapanuiwp

Expert

2022-01-25Added 13 answers

I assume your second ${\mathrm{csc}}^{2}A$ was meant to be ${\mathrm{sec}}^{2}A$ , viz.

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

$=\frac{1}{{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}A}={\mathrm{csc}}^{2}A{\mathrm{sec}}^{2}A$

pripravyf

Expert

2022-01-26Added 12 answers

Let $s=\mathrm{sin}A\text{}\text{and}\text{}c=\mathrm{cos}A$ . 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

$RHS=\frac{{s}^{4}+{c}^{4}+2{s}^{2}{c}^{2}}{{s}^{2}{c}^{2}}=\frac{{({s}^{2}+{c}^{2})}^{2}}{{s}^{2}{c}^{2}}$

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

This gives

But

Since

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