sin left[ cos^{-1} left( frac{3}{5} right) + | vedclass

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Question

(A) Let $\alpha = \cos^{-1} \left( \frac{3}{5} \right)$,then $\cos \alpha = \frac{3}{5}$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have $\sin \alp

Vedclass · Accepted Answer

$\frac{2}{\sqrt{5}}$

Vedclass · Answer

(A) Let $\alpha = \cos^{-1} \left( \frac{3}{5} ight)$,then $\cos \alpha = \frac{3}{5}$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have $\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$. Thus,$\alpha = \sin^{-1} \left( \frac{4}{5} ight)$.Let $\beta = 	an^{-1} 2$,then $	an \beta = 2$. Using the identity $\sin \beta = \frac{	an \beta}{\sqrt{1 + 	an^2 \beta}}$,we get $\sin \beta = \frac{2}{\sqrt{1 + 2^2}} = \frac{2}{\sqrt{5}}$. Thus,$\beta = \sin^{-1} \left( \frac{2}{\sqrt{5}} ight)$.The expression becomes $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$.We have $\sin \alpha = \frac{4}{5}$,$\cos \alpha = \frac{3}{5}$,$\sin \beta = \frac{2}{\sqrt{5}}$,and $\cos \beta = \frac{1}{\sqrt{1 + 2^2}} = \frac{1}{\sqrt{5}}$.Substituting these values: $\sin(\alpha + \beta) = (\frac{4}{5} 	imes \frac{1}{\sqrt{5}}) + (\frac{3}{5} 	imes \frac{2}{\sqrt{5}}) = \frac{4}{5\sqrt{5}} + \frac{6}{5\sqrt{5}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$.

$\sin \left[ \cos^{-1} \left( \frac{3}{5} \right) + \tan^{-1} 2 \right] = $

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