If \cos(\alpha+\beta )=\frac{3}{5},\sin(\alpha-\beta )=\frac{5}{13} and 0<\alpha,\beta <\frac{\pi}{4} then \tan(2\alpha) is equal to :
 

  • Option 1)

    \frac{21}{16}

  • Option 2)

    \frac{63}{16}

  • Option 3)

    \frac{33}{52}

     

  • Option 4)

    \frac{63}{52}

 

Answers (1)

\\\cos (\alpha +\beta )=\frac{3}{5}\\\; \; \; \\\sin (\alpha -\beta )=\frac{5}{13}

0<\alpha , \beta <\frac{4}{4}\; \; \; \; \; \; then\; \; \; \tan(2\alpha )=?

\\\cos(\alpha +\beta )=\frac{3}{5}\\\; \; \; \; \\\tan(\alpha +\beta )=\frac{4}{3}                                         

                                                                             

\\\sin(\alpha -\beta )=\frac{5}{13}\\\; \; \; \; \\\tan(\alpha -\beta )=\frac{5}{12}

\\\alpha+\beta=\tan^{-1}\left ( \frac{4}{3} \right )\\\; \; \; \\\alpha-\beta=\tan^{-1}\left ( \frac{5}{12} \right )

=2\alpha=\tan^{-1}\left ( \frac{4}{3} \right )+\tan^{-1}\left ( \frac{5}{12} \right )

=\tan^{-1}\left ( \frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}} \right )

=\tan^{-1}\left ( \frac{\frac{21}{12}}{\frac{36-20}{36}} \right )=\frac{\frac{21}{12}}{\frac{16}{36}}

=\tan^{-1}\left ( \frac{3\times 21}{16} \right )

=\tan^{-1}\left (\frac{63}{16} \right )

 


Option 1)

\frac{21}{16}

Option 2)

\frac{63}{16}

Option 3)

\frac{33}{52}

 

Option 4)

\frac{63}{52}

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