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3.Trigonometrical Ratios, Functions and Identities
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माना $\cos (\alpha+\beta)=\frac{4}{5}$ और $\sin (\alpha-\beta)=\frac{5}{13},$ जहाँ $0 \leq \alpha, \beta \leq \frac{\pi}{4}$ तो $\tan 2 \alpha$ बराबर है
A
$\frac{{16}}{{63}}$
B
$\frac{{56}}{{33}}$
C
$\frac{{28}}{{33}}$
D
एक पण नहीं
(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
