facas9

2021-08-11

If $\mathrm{tan}A=\mathrm{tan}B$ and $\mathrm{sin}A=m\mathrm{sin}B$, prove that ${\mathrm{cos}}^{2}A=\frac{{m}^{2}-1}{{n}^{2}-1}$

Brittany Patton

We use Trigonometry identities to solve this Question
$\mathrm{tan}A=n\mathrm{tan}B$
$\frac{\mathrm{sin}A}{\mathrm{cos}A}=n\frac{\mathrm{sin}B}{\mathrm{cos}B}$
$\frac{\mathrm{sin}A\mathrm{cos}B}{\mathrm{cos}A\mathrm{sin}B}=n$
$\mathrm{sin}A=m\mathrm{sin}B$
$\frac{\mathrm{sin}A}{\mathrm{sin}B}=m$
${m}^{2}-1={\left(\frac{\mathrm{sin}A}{\mathrm{sin}B}\right)}^{2}-1$
${m}^{2}-1=\frac{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}B}{{\mathrm{sin}}^{2}B}$
${n}^{2}-1={\left(\frac{\mathrm{sin}A\mathrm{cos}B}{\mathrm{cos}A\mathrm{sin}B}\right)}^{2}-1$
$=\frac{{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}B}{{\mathrm{cos}}^{2}A{\mathrm{sin}}^{2}B}-1$
${n}^{2}-1=\frac{{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}B-{\mathrm{cos}}^{2}A{\mathrm{sin}}^{2}B}{{\mathrm{cos}}^{2}A{\mathrm{sin}}^{2}B}$
$\frac{{m}^{2}-1}{{n}^{2}-1}=\frac{\frac{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}B}{{\mathrm{sin}}^{2}B}}{\frac{{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}B-{\mathrm{cos}}^{2}A{\mathrm{sin}}^{2}B}{{\mathrm{cos}}^{2}A{\mathrm{sin}}^{2}B}}$
$\frac{{m}^{2}-1}{{n}^{2}-1}=\frac{{\mathrm{cos}}^{2}A\left({\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}B\right)}{{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}B-{\mathrm{cos}}^{2}A{\mathrm{sin}}^{2}B}$
$=\frac{{\mathrm{cos}}^{2}A\left({\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}B\right)}{{\mathrm{sin}}^{2}A\left(1-{\mathrm{sin}}^{2}B\right)-\left(1-{\mathrm{sin}}^{2}A\right){\mathrm{sin}}^{2}B}$

Jeffrey Jordon