Considering the principal values of inverse t | vedclass

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(B) Let $	heta = \sin^{-1}\left(\frac{2}{\sqrt{13}}ight)$ and $\phi = \cos^{-1}\left(\frac{3}{\sqrt{10}}ight)$.Then $\sin 	heta = \frac{2}{\sqrt

Vedclass · Accepted Answer

$\frac{33}{56}$

Vedclass · Answer

(B) Let $	heta = \sin^{-1}\left(\frac{2}{\sqrt{13}}ight)$ and $\phi = \cos^{-1}\left(\frac{3}{\sqrt{10}}ight)$.Then $\sin 	heta = \frac{2}{\sqrt{13}}$, which implies $	an 	heta = \frac{2}{3}$.Then $\cos \phi = \frac{3}{\sqrt{10}}$, which implies $	an \phi = \frac{1}{3}$.We need to find $	an(2	heta - 2\phi) = \frac{	an 2	heta - 	an 2\phi}{1 + 	an 2	heta 	an 2\phi}$.Using $	an 2\alpha = \frac{2 	an \alpha}{1 - 	an^2 \alpha}$:$	an 2	heta = \frac{2(2/3)}{1 - (2/3)^2} = \frac{4/3}{1 - 4/9} = \frac{4/3}{5/9} = \frac{4}{3} 	imes \frac{9}{5} = \frac{12}{5}$.$	an 2\phi = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{2}{3} 	imes \frac{9}{8} = \frac{3}{4}$.Substituting these values:$	an(2	heta - 2\phi) = \frac{12/5 - 3/4}{1 + (12/5)(3/4)} = \frac{(48-15)/20}{1 + 36/20} = \frac{33/20}{56/20} = \frac{33}{56}$.

Considering the principal values of inverse trigonometric functions, the value of the expression $\tan\left(2 \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to:

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