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3.Trigonometrical Ratios, Functions and Identities
medium
If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$
A
$\frac{{16}}{{63}}$
B
$\frac{{56}}{{33}}$
C
$\frac{{28}}{{33}}$
D
None of these
(AIEEE-2010) (IIT-1979)
Solution
$\cos \,(\alpha \, + \beta ) = \frac{4}{5} \Rightarrow \tan (\alpha \, + \beta ) = \frac{3}{4}$
$\sin \,(\alpha \, – \beta )\, = \frac{5}{{13}} \Rightarrow \tan \,(\alpha \, – \beta ) = \frac{5}{{12}}$
$\tan 2\alpha \, = \tan [(\alpha \, + \beta ) + (\alpha \, – \beta )]$ $\, = \,\frac{{\frac{3}{4} + \frac{5}{{12}}}}{{1 – \frac{3}{4}.\frac{5}{{12}}}} = \frac{{56}}{{33}}$
Standard 11
Mathematics
