# If \tan A=\tan B and \sin A=m\sin B, prove that \cos^2A=\frac{m^2

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

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