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(A) माना $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. तब $\cos \alpha = \frac{1}{5\sqrt{2}}$.चूंकि $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{50}

Vedclass · Accepted Answer

$\frac{3}{29}$

Vedclass · Answer

(A) माना $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. तब $\cos \alpha = \frac{1}{5\sqrt{2}}$.चूंकि $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{50} = \frac{49}{50}$,इसलिए $\sin \alpha = \frac{7}{5\sqrt{2}}$.अतः,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{7/5\sqrt{2}}{1/5\sqrt{2}} = 7$.माना $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$. तब $\sin \beta = \frac{4}{\sqrt{17}}$.चूंकि $\cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{16}{17} = \frac{1}{17}$,इसलिए $\cos \beta = \frac{1}{\sqrt{17}}$.अतः,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.हमें $	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$ का मान ज्ञात करना है।मान रखने पर,$	an(\alpha - \beta) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29}$.

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