If tan ^{-1} sqrt{x^2+x}+sin ^{-1} sqrt{x^2+x | vedclass

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(D) Given the equation: $	an ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ Since $\sin ^{-1} 	heta + \cos ^{-1} 	heta = \frac{\pi}{2}

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(D) Given the equation: $	an ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ Since $\sin ^{-1} 	heta + \cos ^{-1} 	heta = \frac{\pi}{2}$,we can write: $	an ^{-1} \sqrt{x^2+x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x^2+x+1} = \cos ^{-1} \sqrt{x^2+x+1}$ Let $	an ^{-1} \sqrt{x^2+x} = 	heta$. Then $	an 	heta = \sqrt{x^2+x}$. This implies $\cos 	heta = \frac{1}{\sqrt{1 + (x^2+x)}}$. Thus,$\cos ^{-1} \left( \frac{1}{\sqrt{x^2+x+1}} ight) = \cos ^{-1} \sqrt{x^2+x+1}$. Comparing the arguments: $\frac{1}{\sqrt{x^2+x+1}} = \sqrt{x^2+x+1}$ $1 = x^2+x+1$ $x^2+x = 0$ $x(x+1) = 0$ So,$x = 0$ or $x = -1$. Checking the domain of $\sin ^{-1} \sqrt{x^2+x+1}$,we require $0 \le x^2+x+1 \le 1$,which implies $x^2+x \le 0$. For $x=0$,$x^2+x=0$ (valid). For $x=-1$,$x^2+x=0$ (valid). However,$	an ^{-1} \sqrt{x^2+x}$ requires $x^2+x \ge 0$. Thus,$x^2+x=0$ is the only solution,giving $x=0$ or $x=-1$.

If $\tan ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$,then the value of $x$ is

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