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3.Trigonometrical Ratios, Functions and Identities
medium
यदि $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ तब $\cos 2\alpha = $
A
$\sin 2\beta $
B
$\sin 4\beta $
C
$\sin 3\beta $
D
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Solution
(b) $\cos 2\alpha = \frac{{1 – {t^2}}}{{1 + {t^2}}} = \frac{{24}}{{25}}$ {यहाँ $t = \tan \alpha $}
$\sin 2\beta = \frac{{2T}}{{1 + {T^2}}} = \frac{3}{5} \Rightarrow \cos 2\beta = \frac{4}{5}$ {$T = \tan \beta $}
$\therefore \,\,\sin 4\beta = 2\sin 2\beta \cos 2\beta $
$= 2.\frac{3}{5}.\frac{4}{5} = \frac{{24}}{{25}} = \cos 2\alpha $.
Standard 11
Mathematics
