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

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(A) दिया गया है $\sin \beta = \frac{1}{\sqrt{10}}$। चूँकि $0 < \beta < \frac{\pi}{2}$,इसलिए $\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \frac{1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) दिया गया है $\sin \beta = \frac{1}{\sqrt{10}}$। चूँकि $0 अतः,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$।अब,$	an 2\beta = \frac{2 	an \beta}{1 - 	an^2 \beta} = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{2}{3} 	imes \frac{9}{8} = \frac{3}{4}$।अब,$	an(\alpha + 2\beta) = \frac{	an \alpha + 	an 2\beta}{1 - 	an \alpha 	an 2\beta} = \frac{1/7 + 3/4}{1 - (1/7)(3/4)} = \frac{(4+21)/28}{(28-3)/28} = \frac{25}{25} = 1$।चूँकि $0  0$)।अतः,$0 अतः,$2\beta = \frac{\pi}{4} - \alpha$।

यदि $\tan \alpha = \frac{1}{7}$ और $\sin \beta = \frac{1}{\sqrt{10}}$ जहाँ $0 < \alpha, \beta < \frac{\pi}{2}$ है,तो $2\beta$ का मान ज्ञात कीजिए:

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