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

facas9 2021-08-11 Answered
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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Expert Answer

Brittany Patton
Answered 2021-08-12 Author has 27914 answers
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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Answered 2021-12-11 Author has 11052 answers

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