cot^{-1} frac{3}{4} + sin^{-1} frac{5}{13} = | vedclass

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(A) माना $\cot^{-1} \frac{3}{4} = 	heta$,तो $\cot 	heta = \frac{3}{4}$।चूँकि $\cot 	heta = \frac{	ext{आधार}}{	ext{लम्ब}} = \frac{3}{4}$,कर्ण $\sq

Vedclass · Accepted Answer

$\sin^{-1} \frac{63}{65}$

Vedclass · Answer

(A) माना $\cot^{-1} \frac{3}{4} = 	heta$,तो $\cot 	heta = \frac{3}{4}$।चूँकि $\cot 	heta = \frac{	ext{आधार}}{	ext{लम्ब}} = \frac{3}{4}$,कर्ण $\sqrt{3^2 + 4^2} = 5$ होगा।अतः,$\sin 	heta = \frac{4}{5}$,जिसका अर्थ है $	heta = \sin^{-1} \frac{4}{5}$।अब,व्यंजक $\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13}$ हो जाता है।सूत्र $\sin^{-1} x + \sin^{-1} y = \sin^{-1} \left( x \sqrt{1 - y^2} + y \sqrt{1 - x^2} ight)$ का उपयोग करने पर:$= \sin^{-1} \left( \frac{4}{5} \sqrt{1 - \left( \frac{5}{13} ight)^2} + \frac{5}{13} \sqrt{1 - \left( \frac{4}{5} ight)^2} ight)$$= \sin^{-1} \left( \frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5} ight)$$= \sin^{-1} \left( \frac{48}{65} + \frac{15}{65} ight)$$= \sin^{-1} \frac{63}{65}$।

$\cot^{-1} \frac{3}{4} + \sin^{-1} \frac{5}{13} = $

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