The value of sin^{-1} left( frac{12}{13} righ | vedclass

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(D) Let $\alpha = \sin^{-1} \left( \frac{12}{13} \right)$ and $\beta = \sin^{-1} \left( \frac{3}{5} \right)$.Then $\sin \alpha = \frac{12}{13} \implie

Vedclass · Accepted Answer

$\frac{\pi}{2} - \sin^{-1} \left( \frac{56}{65} \right)$

Vedclass · Answer

(D) Let $\alpha = \sin^{-1} \left( \frac{12}{13} ight)$ and $\beta = \sin^{-1} \left( \frac{3}{5} ight)$.Then $\sin \alpha = \frac{12}{13} \implies \cos \alpha = \sqrt{1 - \left( \frac{12}{13} ight)^2} = \frac{5}{13}$.And $\sin \beta = \frac{3}{5} \implies \cos \beta = \sqrt{1 - \left( \frac{3}{5} ight)^2} = \frac{4}{5}$.Using the formula $\sin^{-1} x - \sin^{-1} y = \sin^{-1} (x \sqrt{1 - y^2} - y \sqrt{1 - x^2})$:$\sin^{-1} \left( \frac{12}{13} ight) - \sin^{-1} \left( \frac{3}{5} ight) = \sin^{-1} \left( \frac{12}{13} 	imes \frac{4}{5} - \frac{3}{5} 	imes \frac{5}{13} ight)$$= \sin^{-1} \left( \frac{48}{65} - \frac{15}{65} ight) = \sin^{-1} \left( \frac{33}{65} ight)$.Since $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we have $\sin^{-1} \left( \frac{33}{65} ight) = \frac{\pi}{2} - \cos^{-1} \left( \frac{33}{65} ight)$.Alternatively,using $\cos^{-1} x = \sin^{-1} \sqrt{1 - x^2}$,we have $\sin^{-1} \left( \frac{33}{65} ight) = \cos^{-1} \sqrt{1 - \left( \frac{33}{65} ight)^2} = \cos^{-1} \left( \frac{56}{65} ight) = \frac{\pi}{2} - \sin^{-1} \left( \frac{56}{65} ight)$.Thus,the correct option is $D$.

The value of $\sin^{-1} \left( \frac{12}{13} \right) - \sin^{-1} \left( \frac{3}{5} \right)$ is equal to

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