If tan alpha = frac{1}{7} and sin beta = frac | vedclass

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Question

(a) Since $\sin \beta = \frac{1}{{\sqrt {10} }}$

$\Rightarrow 	an \beta = \frac{1}{3}$

==> $	an 2\beta = \frac{{2	an \beta }}{{1 - {{	an

Vedclass · Accepted Answer

$\frac{\pi }{4} - \alpha $

Vedclass · Answer

(a) Since $\sin \beta = \frac{1}{{\sqrt {10} }}$ $\Rightarrow an \beta = \frac{1}{3}$ ==> $ an 2\beta = \frac{{2 an \beta }}{{1 - {{ an }^2}\beta }} = \frac{3}{4}$ $ herefore an (\alpha + 2\beta ) = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = \frac{{25}}{{25}} = 1$ Now, $0 < \beta < \frac{\pi }{2}$ and $ an 2\beta = \frac{3}{4} > 0$ both ==> $0 < 2\beta < \frac{\pi }{2}$. Again,$0 < \alpha < \frac{\pi }{2}$ and $0 < 2\beta < \frac{\pi }{2}$ both ==> $0 < \alpha + 2\beta < \pi $ Thus, $0 < \alpha + 2\beta < \pi $ and $ an (\alpha + 2\beta ) = 1$ both ==> $\alpha + 2\beta = \frac{\pi }{4} $ $\Rightarrow 2\beta = \frac{\pi }{4} - \alpha $.

If $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{{\sqrt {10} }}\left( {0 < \alpha ,\,\beta < \frac{\pi }{2}} \right)$, then $2\beta $ is equal to

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