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

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(D) माना $\alpha = \sin^{-1} \left( \frac{3}{5} ight)$। तब $\sin \alpha = \frac{3}{5}$,इसलिए $	an \alpha = \frac{3}{\sqrt{5^2 - 3^2}} = \frac{3}{4}

Vedclass · Accepted Answer

$\frac{17}{6}$

Vedclass · Answer

(D) माना $\alpha = \sin^{-1} \left( \frac{3}{5} ight)$। तब $\sin \alpha = \frac{3}{5}$,इसलिए $	an \alpha = \frac{3}{\sqrt{5^2 - 3^2}} = \frac{3}{4}$। अतः,$\alpha = 	an^{-1} \left( \frac{3}{4} ight)$।माना $\beta = \cos^{-1} \left( \frac{3}{\sqrt{13}} ight)$। तब $\cos \beta = \frac{3}{\sqrt{13}}$,इसलिए $	an \beta = \frac{\sqrt{(\sqrt{13})^2 - 3^2}}{3} = \frac{\sqrt{13-9}}{3} = \frac{2}{3}$। अतः,$\beta = 	an^{-1} \left( \frac{2}{3} ight)$।अब,व्यंजक $	an(\alpha + \beta) = 	an \left( 	an^{-1} \frac{3}{4} + 	an^{-1} \frac{2}{3} ight)$ हो जाता है।सूत्र $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ का उपयोग करने पर:$	an(\alpha + \beta) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left( \frac{3}{4} 	imes \frac{2}{3} ight)} = \frac{\frac{9+8}{12}}{1 - \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{6}{12}} = \frac{17}{6}$।

$\tan \left[ \sin^{-1} \left( \frac{3}{5} \right) + \cos^{-1} \left( \frac{3}{\sqrt{13}} \right) \right]$ का मान ज्ञात कीजिए।

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