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Let cos(\alpha +\beta )=\frac{4}{5}  and let sin(\alpha -\beta )=\frac{5}{13} , where 0\leq \alpha , \; \beta \leq \frac{\pi }{4}.  Then tan2\alpha =

  • Option 1)

    \frac{25}{16}

  • Option 2)

    \frac{56}{33}

  • Option 3)

    \frac{19}{12}

  • Option 4)

    \frac{20}{17}

 

Answers (1)

As we leant in

Double Angle Formula -

Double angle formula

- wherein

These are formulae for double angles.

 

 \\ \cos (\alpha + \beta)= \frac{4}{5} \Rightarrow \alpha + \beta = \cos^{-1}\frac{4}{5} ...................(i)

\sin (\alpha - \beta)= \frac{5}{13} \Rightarrow \alpha - \beta = \sin^{-1}\frac{5}{13} ...............(ii)

Adding (1) and (2) 

2 \alpha =cos^{-1}\frac{4}{5}+sin^{-1}\frac{5}{13}

tan \ 2 \alpha =\frac{\tan (\cos^{-1}\frac{4}{5} + \tan (\sin^{-1}\frac{5}{13})}{1- \tan \cos^{1}\frac{4}{5} \tan \sin ^{-1}\frac{3}{13}}

= \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}} = \frac{56}{33}

 


Option 1)

\frac{25}{16}

This option is incorrect.

Option 2)

\frac{56}{33}

This option is correct.

Option 3)

\frac{19}{12}

This option is incorrect.

Option 4)

\frac{20}{17}

This option is incorrect

Posted by

Vakul

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