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(B) Given $	an \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$.Since $\beta$ is in the first quadrant,$\cos \beta = \sqrt{1 - \sin^2 \be

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Given $	an \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$.Since $\beta$ is in the first quadrant,$\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.Thus,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$.We need to find $\alpha + 2\beta = 	an^{-1}(\frac{1}{7}) + 2	an^{-1}(\frac{1}{3})$.Using the formula $2	an^{-1} x = 	an^{-1}(\frac{2x}{1-x^2})$,we get:$2	an^{-1}(\frac{1}{3}) = 	an^{-1}(\frac{2(1/3)}{1-(1/3)^2}) = 	an^{-1}(\frac{2/3}{8/9}) = 	an^{-1}(\frac{2}{3} 	imes \frac{9}{8}) = 	an^{-1}(\frac{3}{4})$.Now,$\alpha + 2\beta = 	an^{-1}(\frac{1}{7}) + 	an^{-1}(\frac{3}{4})$.Using $	an^{-1} x + 	an^{-1} y = 	an^{-1}(\frac{x+y}{1-xy})$:$\alpha + 2\beta = 	an^{-1}(\frac{1/7 + 3/4}{1 - (1/7)(3/4)}) = 	an^{-1}(\frac{4/28 + 21/28}{1 - 3/28}) = 	an^{-1}(\frac{25/28}{25/28}) = 	an^{-1}(1) = 45^{\circ}$.

If $\alpha$ and $\beta$ are angles in the first quadrant such that $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$,then $\alpha + 2\beta =$ (in $^{\circ}$)

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