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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$(S1) :$ જો $P(A) =0$ હોય, તો $A =\emptyset$
$(S2) :$ જો $P ( A )=1$ હોય, તો $A =\Omega$
તો