Let f(x) = (sin(tan^{-1} x) + sin(cot^{-1} x) | vedclass

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(A) Let $	an^{-1} x = 	heta$. Then $\sin(	an^{-1} x) = \frac{x}{\sqrt{1+x^2}}$ and $\sin(\cot^{-1} x) = \frac{1}{\sqrt{1+x^2}}$.$f(x) = (\frac{x+1}

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$\frac{5\pi}{6}$

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(A) Let $	an^{-1} x = 	heta$. Then $\sin(	an^{-1} x) = \frac{x}{\sqrt{1+x^2}}$ and $\sin(\cot^{-1} x) = \frac{1}{\sqrt{1+x^2}}$.$f(x) = (\frac{x+1}{\sqrt{1+x^2}})^2 - 1 = \frac{x^2+1+2x}{1+x^2} - 1 = \frac{2x}{1+x^2}$.Given $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(\sin^{-1}(\frac{2x}{1+x^2}))$.For $|x| > 1$,$\sin^{-1}(\frac{2x}{1+x^2}) = \pi - 2	an^{-1} x$ if $x > 1$ and $-\pi - 2	an^{-1} x$ if $x Thus,$\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(\pi - 2	an^{-1} x) = -\frac{1}{1+x^2}$ for $x > 1$.And $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(-\pi - 2	an^{-1} x) = -\frac{1}{1+x^2}$ for $x Integrating,$y = -	an^{-1} x + C_1$ for $x > 1$ and $y = -	an^{-1} x + C_2$ for $x Given $y(\sqrt{3}) = \frac{\pi}{6} \Rightarrow -\frac{\pi}{3} + C_1 = \frac{\pi}{6} \Rightarrow C_1 = \frac{\pi}{2}$.Since the function is not defined at $x=0$,$C_2$ cannot be determined from $C_1$ without further information. Assuming the function is continuous or symmetric,if $C_2 = C_1 = \frac{\pi}{2}$,then $y(-\sqrt{3}) = -	an^{-1}(-\sqrt{3}) + \frac{\pi}{2} = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}$.

Let $f(x) = (\sin(\tan^{-1} x) + \sin(\cot^{-1} x))^2 - 1$ for $|x| > 1$. If $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(\sin^{-1}(f(x)))$ and $y(\sqrt{3}) = \frac{\pi}{6}$,then $y(-\sqrt{3})$ is equal to

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