If tan A = frac{1}{sqrt{x(x^2+x+1)}}, tan B = | vedclass

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(A) We are given $	an A = \frac{1}{\sqrt{x(x^2+x+1)}}$ and $	an B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$.Using the formula $	an(A+B) = \frac{	an A + \

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$C$

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(A) We are given $	an A = \frac{1}{\sqrt{x(x^2+x+1)}}$ and $	an B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$:$	an(A+B) = \frac{\frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \frac{1}{\sqrt{x(x^2+x+1)}} \cdot \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}$$= \frac{\frac{1 + x}{\sqrt{x(x^2+x+1)}}}{1 - \frac{\sqrt{x}}{\sqrt{x(x^2+x+1)^2}}} = \frac{\frac{1+x}{\sqrt{x(x^2+x+1)}}}{1 - \frac{1}{x^2+x+1}}$$= \frac{(1+x) \sqrt{x^2+x+1}}{\sqrt{x}(x^2+x+1 - 1)} = \frac{(1+x) \sqrt{x^2+x+1}}{\sqrt{x}(x^2+x)}$$= \frac{(1+x) \sqrt{x^2+x+1}}{\sqrt{x} \cdot x(x+1)} = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} = \sqrt{\frac{x^2+x+1}{x^3}}$$= \sqrt{x^{-1} + x^{-2} + x^{-3}} = 	an C$.Since $0 < A, B, C < \frac{\pi}{2}$,we have $A+B = C$.

If $\tan A = \frac{1}{\sqrt{x(x^2+x+1)}}, \tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C = (x^{-3}+x^{-2}+x^{-1})^{\frac{1}{2}}$,where $0 < A, B, C < \frac{\pi}{2}$,then $A+B$ is equal to:

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