If sin ^{-1} x-cos ^{-1} 2 x=sin ^{-1}left(fr | vedclass

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(B) Given equation: $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$.Since $\sin ^{-1

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$\frac{\pi}{4}$

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(B) Given equation: $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}ight)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}ight)$.Since $\sin ^{-1}\left(\frac{\sqrt{3}}{2}ight)=\frac{\pi}{3}$ and $\cos ^{-1}\left(\frac{\sqrt{3}}{2}ight)=\frac{\pi}{6}$, we have $\sin ^{-1} x-\cos ^{-1} 2 x=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$.Thus, $\sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} 2 x$.Taking sine on both sides: $x=\sin\left(\frac{\pi}{6}+\cos ^{-1} 2 xight) = \sin\left(\frac{\pi}{6}ight)\cos\left(\cos ^{-1} 2 xight) + \cos\left(\frac{\pi}{6}ight)\sin\left(\cos ^{-1} 2 xight)$.$x = \frac{1}{2}(2x) + \frac{\sqrt{3}}{2}\sqrt{1-(2x)^2} = x + \frac{\sqrt{3}}{2}\sqrt{1-4x^2}$.This implies $\frac{\sqrt{3}}{2}\sqrt{1-4x^2} = 0$, so $1-4x^2=0$, which gives $x = \frac{1}{2}$ (as $x = -\frac{1}{2}$ does not satisfy the original equation).Now, calculate $	an ^{-1} x+	an ^{-1}\left(\frac{x}{x+1}ight)$ for $x=\frac{1}{2}$:$	an ^{-1}\left(\frac{1}{2}ight)+	an ^{-1}\left(\frac{1/2}{1/2+1}ight) = 	an ^{-1}\left(\frac{1}{2}ight)+	an ^{-1}\left(\frac{1}{3}ight)$.Using the formula $	an ^{-1} a + 	an ^{-1} b = 	an ^{-1}\left(\frac{a+b}{1-ab}ight)$:$	an ^{-1}\left(\frac{1/2+1/3}{1-(1/2)(1/3)}ight) = 	an ^{-1}\left(\frac{5/6}{5/6}ight) = 	an ^{-1}(1) = \frac{\pi}{4}$.

If $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$, then $\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=$

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