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

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Question

(A) Let $\cot^{-1} \frac{3}{4} = 	heta$,then $\cot 	heta = \frac{3}{4}$.Since $\cot 	heta = \frac{	ext{base}}{	ext{perpendicular}} = \frac{3}{4}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $\cot^{-1} \frac{3}{4} = 	heta$,then $\cot 	heta = \frac{3}{4}$.Since $\cot 	heta = \frac{	ext{base}}{	ext{perpendicular}} = \frac{3}{4}$,the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.Therefore,$\sin 	heta = \frac{4}{5}$,which implies $	heta = \sin^{-1} \frac{4}{5}$.Now,the expression becomes $\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13}$.Using the formula $\sin^{-1} x + \sin^{-1} y = \sin^{-1} \left( x \sqrt{1 - y^2} + y \sqrt{1 - x^2} ight)$:$= \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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