यदि tan alpha = frac{1}{7} तथा sin beta | vedclass

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Question

चूँकि $\sin \beta  = \frac{1}{{\sqrt {10} }} \Rightarrow 	an \beta  = \frac{1}{3}$

$\Rightarrow$ $	an 2\beta  = \frac{{2	an

Vedclass · Accepted Answer

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

Vedclass · Answer

चूँकि $\sin \beta  = \frac{1}{{\sqrt {10} }} \Rightarrow 	an \beta  = \frac{1}{3}$

$\Rightarrow$ $	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$

अब $0 < \beta  < \frac{\pi }{2}$ एवं $	an 2\beta  = \frac{3}{4} > 0$ दोनों

$\Rightarrow$  $0 < 2\beta  < \frac{\pi }{2}$.

पुन: $0 < \alpha  < \frac{\pi }{2}$ एवं $0 < 2\beta  < \frac{\pi }{2}$ दोनों 

$\Rightarrow$  $0 < \alpha  + 2\beta  < \pi $

अत: $0 < \alpha  + 2\beta  < \pi $ एवं $	an (\alpha  + 2\beta ) = 1$ दोनों     

$\Rightarrow$  $\alpha  + 2\beta  = \frac{\pi }{4} $

$\Rightarrow 2\beta  = \frac{\pi }{4} - \alpha $.

यदि $\tan \alpha = \frac{1}{7}$ तथा $\sin \beta = \frac{1}{{\sqrt {10} }}\left( {0 < \alpha ,\,\beta < \frac{\pi }{2}} \right)$, तब $2\beta $ बराबर है

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