MCQ
If $\cos (\alpha + \beta ) = \frac{4}{5},\sin (\alpha - \beta ) = \frac{5}{{13}}$ and $\alpha ,\beta $ lie between $0$ and $\frac{\pi }{4},$ then $\tan 2\alpha = $
- A$\frac{{16}}{{63}}$
- ✓$\frac{{56}}{{33}}$
- C$\frac{{28}}{{33}}$
- DNone of these
and $\sin \,(\alpha - \beta ) = \frac{5}{{13}}$
$ \Rightarrow \,\,\sin \,(\alpha + \beta ) = \frac{3}{5}$
and $\cos \,(\alpha - \beta ) = \frac{{12}}{{13}}$
$ \Rightarrow \,\,2\alpha = {\sin ^{ - 1}}\frac{3}{5} + {\sin ^{ - 1}}\frac{5}{{13}}$
$ = {\sin ^{ - 1}}\left[ {\frac{3}{5}\sqrt {1 - \frac{{25}}{{169}}} + \frac{5}{{13}}\sqrt {1 - \frac{9}{{25}}} } \right]$
$ \Rightarrow \,\,2\alpha = {\sin ^{ - 1}}\,\left( {\frac{{56}}{{65}}} \right)\,$
$\Rightarrow \,\sin \,2\alpha = \frac{{56}}{{65}}$
Now, $\tan \,2\alpha = \frac{{\sin \,2\alpha }}{{\cos \,2\alpha }} $
$= \frac{{56/65}}{{33/65}} = \frac{{56}}{{33}}$.
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